Molecular Theory of gases - Physics II _ For HSC Part 2 / XII / Class 12 (Science Group) - Section C - ERQs (Long Answered Questions)

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Unit 15: Molecular Theory of gases
Physics II -
For HSC Part 2 / XII / Class 12 (Science Group)

SECTION C
ERQs (Long Answered Questions)

Q.1: What is temperature? Explain the scales of temperature in detail.
Ans: TEMPERATURE:
It is the measurement of hotness or coldness in a body. It determines the direction of flow of heat.
Definition:
"Temperature is a measure of the average translational kinetic energy of the molecules of body."
Unit:
The SI unit of temperature is Kelvin.

Measurement Of Temperature:
Temperature is measured by thermometer. It can be determine by measuring the physical properties, as properties changes with the temperature.

Scales of Temperature OR Types Of Thermometer:
There are three scales of temperature which are commonly used these days.
i. Centigrade or Celsius scale
ii. Fahrenheit Scale
iii. Kelvin or Absolute Scale

i) Centigrade or Celsius Scale (Anders Celsius):
In the Celsius scale the freezing point of water or Melting point of ice is marked as 0 °C and boiling point of water is marked as 100 °C. The interval between these two points is divided into hundred equal parts. Each part thus represents one degree centigrade or Celsius and is represented by °C.

ii) Fahrenheit Scale (Daniel Fahrenheit):
In Fahrenheit scale the freezing point of water is marked 32 °F and boiling point of water is marked as 212°F respectively and the interval between the two fixed points (freezing & boiling points) is divided into 180 equal parts. Each part is called one degree Fahrenheit and is represented by 1°F.

(iii) Kelvin Scale (James Lord Kelvin):
In this scale, the melting point (freezing point) of ice is marked as 273 K and the boiling point of water is marked as 373 K. The interval between two fixed points (freezing & boiling points) is divided into 100 equal parts. The temperature is given in units called Kelvin instead of degrees and is represented by K. The lowest temperature is 0 K known as absolute zero.

Empirical Formulae:
In order to derive empirical formulae among Centigrade, Fahrenheit and Kelvin scales, let the three thermometer placed in a bath tub and the mercury in each thermometer rises to the same level. Thus we arrive at the relation.

Relation Between Different Temperature Scales:
Celsius (Centigrade) Scale, Kelvin Scale & Fahrenheit Scale:

• Centigrade temperature to Kelvin temperature by simply adding 273 to the centigrade temperature i.e.
  T_K = T_°C + 273
  
  
• Kelvin to Celsius
  T_°C = T_K - 273
  
  
• Fahrenheit scale To Celsius Scale:
  T_°C = (⁵/₉ x T °F) - 32
  
  
• Celsius To Fahrenheit scale:
  T_°F = (⁹/₅ x T °C) + 32
  
  
• Scale Division:
  * The scale divisions on the Kelvin scale are equal in size to the divisions on Celsius scale. i.e. 1K= 1 °C
  * 1 °C = 1.8 °F
  * Fahrenheit and Celsius scales coincide at - 40 °C OR - 40 °F
  * Fahrenheit and Kelvin scales coincide at 574.25° F or 574.25K

Q.2: Define and explain Boyle's law, Charles's and Avogadro's law.
Ans: (a) BOYLE's LAW:
In 1662 Robert Boyle proposed gases law about the relationship between volume and pressure of an ideal at constant temperature and mass is known as Boyle's law.

Definition (Statement):
Boyle's law states that:

"Volume V of any given mass of a gas is inversely proportional to the pressure P, provided the temperature T of the gas remains constant."

Explanation:
Mathematically:
If V is the volume of the gas and P is its pressure, then Boyle's law can be written as:

V ∝ ¹ /_P (at constant temperature)
V = ^K/_P
OR
PV = K

Where K is a constant and depends upon sample of gas.
For change in state from initial to final, Boyle's law can be written as:

P₁ V₁ = P₂ V₂ = Constant

Real gases obey Boyle's law at low pressure.

Graphically:
We can also represent Boyle's law on a graph, as:

• The graph plotted between P and V at constant temperature is a curve called hyperbola showing the inverse relation between them for two different states.
• While the second graph of P plotted against ¹ /_v is a straight line passing through the origin, showing direct proportionality.

The relation between pressure and volume & pressure and inverse of volume respectively

Relation with the mass (m) of a gas:
The inverse relation of volume and pressure of a gas remain stable until the mass of the gas remains constant. If, however, the mass of gas varies, then the product of PV will be directly proportional to the mass of a gas:

PV ∝ m
PV = Constant x m
OR
PV / m = Constant

For different state of gases

P₁ V₁/ m₁ = P₂ V₂/m₂ = Constant

(b) CHARLE'S LAW:
In 1787 French scientist J. Charles proposed his law to explain the relationship between volume and absolute temperature of a given mass of a gas, at the pressure constant is known as Charle's Law.

Definition (Statement): This law states that:

"The volume V of any given mass of a gas is directly proportional to the absolute temperature T at constant pressure P".

Explanation:
Mathematically:
Charles law can be written as:

V ∝ T (at constant pressure)
V = KT
OR
V / T = K

Where K is constant and depends upon sample of a gas.
If V₁ and V₂ are the volumes of the gas at temperature T₁ and T₂ respectively then for two different states, For change in state from initial to final, Charles's law is represented as:

V₁/T₁ = V₂/T₂ = constant

Real gases obey Charle's law at high temperature.

Graphically:

• From the graph at 0°C the gas still has a volumeV₀.
• The graph between volume and temperature is a straight line, which shows that volume increase with the increase in temperature.
• If the this line is extrapolated (projected) backward, it meets the temperature axis at -237 °C. indicating volume of a gas is zero.
• Thus (-273°C) is the lowest possible temperature that a gas can achieve and is called Absolute Zero or 0 Kelvin.

Relation between temperature and volume

Relation with the mass (m) of a gas:
The direct relation of volume and temperature of a gas remain stable until the mass of the gas remains constant. If, however, the mass of gas varies, then the ratio will be directly proportional to the mass of a gas:

V/T ∝ m
V/T = Constant x m
OR
V / mT = Constant

For different state of gases:

V₁/ m₁T₁ = V₂/m₂T₂ = Constant

(c) AVOGARDRO'S LAW:
In 1811, Italian scientist Amedeo Avogadro suggested his hypothesis regarding the relationship between volume and number of molecules of a gas. This hypothesis now called Avogadro's law.

Definition (Statement):
Avogadro's law states that:

"equal volume of all gases contains the same number of molecules at the same temperature and pressure".

Explanation:
Mathematically:
Thus, the volume of a gas is directly proportional to number of moles of the gas at constant temperature and pressure. In symbol, we can write as:

V ∝ n (at constant temperature and pressure)
V = Kn
OR
V/n = K

Where K is constant and depends upon a sample of a gas, V is volume and n is the number of moles of a gas.
It can be written as:

V₁/n₁ = V₂/n₂ = Constant

When V₁ and V₂ are volumes of gas and n1₁ and n₂ are amount of gas.

Thus 1 dm³ (or cm³, m3³) of oxygen contains the same number of molecules as 1 dm³ or 1 cm³, 1m³ etc of hydrogen or of any other gas, provided the volumes are measured under the same conditions of temperature and pressure.

Example:
Blowing up a balloon is an example of Avogadro's law, because as we blow more molecules of air into the balloon it expands.

Q.3: Derive general gas law by making use of gas laws.
Ans: GENERAL GAS LAW:

An interrelation among the physical quantities e g. pressure (P), volume (V), temperature (T) and mass (m) of a given sample of gas which determine the state of a gas is termed as "equation of state" of gas or General gas law.

In order to derive general gas law, we make use of Boyle's law, Charles's law and Avogadro's law.
According to Boyle's law:

V ∝ 1/P (when n number of moles and temperature T are kept constant)

According to Charles's law:

V ∝ T (when n and pressure P are kept constant)

According to Avogadro's law:

V ∝ n (when T and P are kept constant)

Lets Consider that none of the variable are to be kept constant, then by the combination of all the above three relationships, we can write as:

Where R is constant of proportionally and is called General gas constant or universal gas constant and does not depend on the quantity of gas in the sample. It is denoted by R.
Above equation is written as:

PV = nRT

S.I unit:
If P is measured in Nm^-2 V in m³ and T in Kelvin then the volume of universal gas constant is R = 8.314 J mol^-1 k^-1

Q.4: Describe the molecular movement causes the pressure exerted by gas, derive pressure equation?
Ans: PRESSURE OF GAS:
The pressure exerted by a gas is merely the momentum transferred to the walls of the container per second per unit area due to the continuous collisions of molecules of the gas.

PRESSURE EQUATION:
To Calculate The Pressure Of An Ideal Gas From Kinetic Theory:

• Let us consider a cubical container having side length L whose walls are perfectly elastic contains N number of molecules each of mass m.
• A molecule which has a velocity V₁ can be resolved into three rectangular components V₁x, V₁y and V₁z parallel to three co-ordinates axes x, y and z.
• A molecule which collides with the face ABCDA of the cubical container, it will rebound elastically in opposite direction, such that component of the velocity V₁ is reversed, and other components remain unaffected.

Now Consider a single molecule of mass m moving with velocity V₁ parallel to x-axis. It moves back and forth, colliding at reguarl intervals with the ends of the box (cubical container) and thereby contributing to the pressure of the gas.
The molecule from the left face of the cube moves with the velocity V₁x and collide with the right face of the cube and rebounds with the velocity -V₁x
Therefore,
Initial momentum of molecule (before collision) = P_i = mV₁x
Final momentum of molecule (after collision) = P_f = m (-V₁x) = - mV₁x
and

Change in momentum = △P = P_i — P_f
= mV₁x — (—mV₁x) = mV₁x + mV₁x
= 2mV₁x .........(1)

After recoil the molecule travels to opposite face and collides with it, rebounds and travels back to the face ABCDA after covering a distance 2L. i.e.

S = 2L
But S = vt
2L = v₁x x t
t = 2L / v₁x
The time t between two successive collisions with face ABCDA = △t
so t = △t
Or △t = 2L / v₁x .........(2)

Now we can find the force exerted by one molecule on face ABCDA, using Newton's 2^nd law of motion. This says that the rate of change of momentum of the molecule is equal to force applied by the wall. According to Newton's 3^rd law of motion, force F₁ exerted the molecule on face ABCDA is equal but opposite.

Similarly, the forces due to all other molecules can be determined. Thus, the total x-directed F due to N number of molecules of the gas moving with velocities V₁, V₂, V₃.........V_n is:

F = F₁ + F₂ + F₃ + ........... + F_n

Q.5: Interpret mathematically that temperature is a measure of average translational K.E of the molecules of a gas. OR
Prove that the average translatory kinetic energyof gas molecule is directly proportional to the Absolute temperature of the gas. OR
Prove that T ∝ (K.E)_ave OR
On the basis of KMT of gases show that ¹/₂ mV² = ³/₂ KT
Ans: THE RELATION BETWEEN KINETIC ENERGY OF MOLECULE AND ABDOLUTE TEMPERATURE:
Consider a gas in a cylinder having density "𝝆", Pressure "P", Volume "V", moles "n" and total number of molecules "N" for such a gas using kinetic theory of gases can be written as:

Note: Average Transitional Kinetic Energy is also called Kinetic Energy per molecule of the gas i.e.:

K.E per molecule = K.E_avg = ³/₂ kT

While the Kinetic Energy of per mole of a gas is:

K.E per mole = ³/₂ RT

More Long Answered Questions

Q.6: Write down the relation of different scales in detail?
Ans: Relations Between Different Temperature Scales:
Relation between Celsius and Fahrenheit Scales:
Let us consider that both, a Celsius and a Fahrenheit, thermometers are dipped in the same water bath placed at room temperature. The final and initial points in Celsius thermometer are labelled as A and B, and those are in the Fahrenheit thermometer as A' and B' respectively, whereas the mercury levels in both are labelled as C and C'.
The general relation between two scales of temperature can be written as:

Relation between Kelvin and Fahrenheit Scales:

Relation between Celsius and Kelvin Scales:

• T °C = TK — 273
• TK = T °C + 273

Q.7: Define and explain triple point of a water?
Ans: The Triple Point of Water:
Definition:

The specific temperature and pressure, where all three states or phases (vapour or gas, liquid and solid) of water are coexist in thermodynamic equilibrium is called Triple Point of Water. This means that at the triple point, ice, liquid water, and water vapor can exist together without any phase changing to another.

Value and Conditions:

• The particular temperature of triple point of water at which pure water, pure ice, and pure water vapour can coexist in a stable equilibrium is precisely 273.16 K (0.01°C) and 32.02°F 
• While the pressure is 4.58 mm of mercury and 611.73 Pascal (N/m²) and is used to calibrate thermometer.
•  These conditions are used to define the Kelvin temperature scale.
• The temperature at which Triple point of water has been set by international agreement to be T₃ = 273.16 K. In T3₃ the subscript 3 means "Triple Point" this agreement also sets the size of the Kelvin as ¹/_273.16 of the difference between the Triple point temperature of water and absolute zero.

Graph OR Phase Diagrams:

• In a graph or phase diagram, the triple point is represented as the point (at the intersection) where the lines separating the solid, liquid, and gas phases meet.
• This graph visually demonstrates the relationship between temperature and pressure for the different phases of water.
• In each of the three regions of the diagram, the substance is in a single state (or phase).
• The dark lines that act as the boundary between those regions represent the conditions under which the two phases are in equilibrium.

Triple Point Cell:

• A triple point cell is a device used to determine the triple point of a substance like water.
• Inside the cell conditions are maintained, so that solid ice, liquid water and water vapour co-exist in thermal equilibrium.
• The bulb of the constant volume gas thermometer is inserted into the well of the cell to measure the temperature at this fixed point.

Q.8: What is kinetic molecular theory of a gas? State the basic postulates of Kinetic theory of gas? Also write down example of KMT
Ans: KINETIC MOLECULAR THEORY:
The Kinetic Theory of Gases (KMT) is a microscopic model that explains the behavior of ideal gases by considering their molecules. It provides a link between the macroscopic properties of gases, such as pressure and temperature, and the microscopic characteristics of molecules, such as their speed and collisions. This theory helps us understand the underlying reasons for the observed properties of gases and forms the basis for many fundamental gas laws.

BASIC POSTULATES OF KINETIC MOLECULAR THEORY:
The basic postulates of Kinetic Theory of gases are as under :

1. MOLECULES:
   A gas contains a very large number of particles called molecules. Depending on the gases each molecule consists of an atom or a group of atoms.
2. VOLUME OF A GAS:
   A finite volume of a gas consists of very large number of molecules. This assumption is justified by experiments. At standard conditions there are 3 x 10²⁵ molecules per cubic meter.
3. SIZE OF MOLECULES:
   The size of the molecules is much smaller than separation between molecules; it is about 3 x 10^-10 m.
4. COLLISION:
   The molecules move in all directions with various speeds collide elastically with one another and with the walls of the container.
5. FORCES BETWEEN MOLECULES:
   The molecules exert no forces on one another except during collisions. In the absence of the external forces, they move freely in straight lines.
6. LAWS OF MECHANICS:
   Laws of mechanics are assumed to be applicable to the motion of molecules.

EXAMPLE:
Popcorn is a fun way to learn about the kinetic molecular theory of gases, the phase change of water from a liquid to a gas. When heated, the water inside turns to steam, making the kernel pop and expand up to 50 times to its size.

Particles Physics - Physics For HSC Part 2 / XII / Class 12 (Science Group) - Section A - Multiple Choice Questions, Fill In The Blanks, Key Formulae & Concept Map

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Unit 28: Particles Physics
Multiple Choice Questions

SECTION A
Multiple Choice Questions (MCQs)

Choose the correct answer:
1. The Standard Model classifies elementary particles into two main groups:
a) Baryons and Leptons
b) Fermions and Bosons ✔
c) Quarks and Gluons
d) Hadrons and Mesons

2. The following is NOT a flavor of quark:
a) Up
b) Down
c) Electron ✔
d) Top

3. The charge of an up quark is:
a) +1/2 e
b) -1/2 e
c) +2/3 e ✔
d) -2/3 e

4. The concept of 'quark confinement' implies that:
a) Quarks cannot exist as free particles outside of hadrons. ✔
b) Quarks are always found in pairs with opposite charges.
c) Quarks have a strong affiinty for gluons.
d) Quarks are the fundamental building blocks of all matter.

5. The primary role of the Higgs boson in the Standard Model is:
a) Mediating electromagnetic interactions
b) Providing mass to other particles ✔
c) Transmitting the strong nuclear force
d) Creating dark matter

6. According to the Standard Model, the term "color' refers to:
a) Visible light spectrum
b) Charge property of quarks ✔
c) Mass of particles
d) Spin of particles

7. A particle made up of a quark and an antiquark is called:
a) Lepton
b) Baryon ✔
c) Meson
d) Neutrino

8. An elementary particle that feels all three fundamental forces (electromagnetic, weak, and strong nuclear forces) is:
a) Lepton
b) Quark ✔
c) Electron
d) Neutrino

9. The primary function of a Geiger-Mtiller counter in particle physics is to:
a) Measure the velocity of particles
b) Detect and count ionizing radiation ✔
c) Create antimatter particles
d) Generate magnetic fields

10. In a Geiger-Muller counter. the gas commonly used to detect ionizing radiation is:
a) Oxygen
b) Neon
c) Argon ✔
d) Helium
Note: (Although all the three are used in Geiger-Muller counter, but:
Argon: Optimal ionization properties, stable, widely used.
Neon and Helium: Less efficient ionization, higher ionization potential, so less practical for standard detection.)

Molecular Theory of gases - Physics II- For HSC Part 2 / XII / Class 12 (Science Group) - Section D: Numerical & Worked Exmples

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Unit 15: Molecular Theory of gases
Physics II
For HSC Part 2 / XII / Class 12 (Science Group)

SECTION D
Numerical

15.1. The freezing point of mercury is —39 °C. Convert it into °F and the comfort level temperature of 20° into Kelvin. (Ans: —38.2 °F, 293 K)

15.2. The boiling point of liquid nitrogen is -321°F. Change it into equivalent Kelvin temperature. (Ans: 77K)

15.3. Calculate the volume occupied by a gram-mole of a gas at 0 °C and a pressure of 1.0 atmosphere. (Ans: 22.4 liters/mole)
DATA:
Given:
Number of moles = n = 1 gram mole
Temperature = 0 °C = 0 + 273 = 273 K
Pressure = P = 1 atm = 1.01 x 10⁵ N/m²
Universal gas Constant = R = 8.314 J mol^-1 K^-1

Required:
Volume = V = ?

Formula:
PV = nRT

Solution:

PV = nRT
1.01 x 10⁵ x V = 1 x 8.314 x 273

V = 0.2247 m³
V = 0.2247 x 1000 = 22.47 L /mole

Result: The volume occupied is 22.47 liters / mole Answer

15.4. An air storage tank whose volume is 112 liters contain 3 kg of air at a pressure of 18 atmospheres. How much air would have to be forced into the tank to increase the pressure to 21 atmospheres, assuming no change in temperature? (Ans: 0.5 kg)
DATA:
Given:
Volume = V₁ = V₂ = 112 liters
Mass of air = m₁ = 3 kg
Pressure = P₁ = 18 atm
Pressure = P₂ = 21 atm
Temperature = T (There is no change in temperature)

Required:
Mass of air = m₂ = ?

Formula:

• Δm = m₁ - m₂
• PV = nRT

Solution:

Δm = m₁ - m₂
Δm = 3.5 - 3 = 0.5 kg

Result: 0.5 kg air would have to be forced. Answer

15.5. A balloon contains 0.04 m³ of air at a pressure of 120 KPa. Calculate the pressure required to reduce its volume to 0.025 m³ at constant temperature. (Ans: 1.9 x 10⁵ Pa)

15.6. The molar mass of nitrogen gas N₂ is 28 gmol^-1. For 100 g of nitrogen, calculate.
(a) the number of moles. (Ans: (a) 3.57 mole)
(b) the volume occupied at room temperature (20 °C) and pressure of 1.01 x 10⁵ Pa. (Ans: (b) 0.086 m³ or 86 dm³)
DATA:
Given:
Molar mass of Nitrogen = M = 28 gmol^-1
Mass of Nitrogen= m = 100 g
Temperature = T = 20 °C = 20 + 273 = 293 K
Pressure = P = 1.01 x 10⁵ Pa
Universal gas Constant = R = 8.314 J mol^-1 K^-1

Required:
(a) Number of moles = n = ?
(b) Volume = V = ?

Formula:

1. Number of moles = mass / Molar mass
2. PV = nRT

Solution:

Result: (a) Number of moles are 3.57 moles
(b) The volume occupied is 0.086 m³ or 86 dm³

15.7. A sample of a gas contains 3.0 x 10²⁴ atoms. Calculate the volume of the gas at a temperature of 300 K and a pressure of 120 KPa. (Ans: 0.104 m³)
DATA:
Given:
Number of Atoms = N = 3.0 x 10²⁴ atoms
Avogadro's Number = N_A = 6.022 x 10²³
Temperature = T = 00 K
Pressure = P = 120 KPa = 120 x 10³ Pa

Required:
Volume of the gas = V = ?

Formula:

• Number of moles = Number of atoms / Avogadro's number
• PV = nRT

Solution:

Result: The volume of gas is 0.104 dm³ Answer

15.8. Calculate the root mean square speed of hydrogen molecules at 0°C and 1.0 atm pressure. Assuming hydrogen to be an ideal gas. The density of hydrogen is 8.99 x10^-2 kg/m³. (Ans: 1835.86 ms^-1)
DATA:
Given:
Temperature = T = 0°C = 0 + 273 = 273 K
Pressure = 1 atm = 1.01 x 10⁵ Pa
Density = = 8.99 x10^-2 kg/m³

Required:
Root mean square speed = V_rms = ?

Formula:
P = ¹/₃ 𝝆 V²

Solution:

Result: The root mean square speed is 1835.8 m/s Answer

15.9. Calculate the root mean square speed of hydrogen molecule at 500 K (mass of proton = 1.67 x 10^-27 kg and K = 1.38 x 10^-23 J/molecule.-K) (Ans: 2489.49ms^-I)
DATA:
Given:
Temperature = T = 500 K
Mass of proton = 1.67 x 10^-27 kg
Mass of hydrogen molecule = H₂ = m = 2 x 1.67 x 10^-27 = 3.34 x 10^-27 kg
Boltzmann constant = K = 1.38 x 10^-23 J/molecule.k
Pressure = 1 atm = 1.01 x 10⁵ Pa

Required:
Root mean square speed = V_rms = ?

Formula:
V² = 3KT / m

Solution:

Result: Root mean square speed is 2489.50 m/s Answer

15.10. (a) Determine the average value of the Kinetic energy of the particles of an ideal gas at 10 °C and at 40 °C. (Ans: 5.86x10^-21 J, 6.48x10^-21 J)
(b) What is the Kinetic energy per mole of an ideal gas at these temperatures? (Ans: 3526.57 J, 3901 J)
DATA:
Given:
Temperature = T = 10 °C = 10 + 273 = 283 K
Temperature = T = 40 °C = 40 + 273 = 313 K
Boltzmann constant = K = 1.38 x 10^-23 J/molecule.k
Number of moles = n = 1 mole
Universal gas Constant = R = 8.314 J mol^-1 K^-1

Required:
(a) Average value of Kinetic Energy = K.E_avg = ?
(b) Kinetic energy per mole of ideal gas = K.E = ?
Formula:

• (a) K.E_avg = ³/₂ KT
• K.E (per mole) = ³/₂ nRT

Solution:

Result:
(a-i) 5.858 x 10^-21
(a-ii) 6.479 x 10^-21

(b-i) 3529.3 J
(b-ii) 3903.4 J Answer

Worked Examples

Numerical From Self Assessment Questions

1. Convert each of the following temperature from the centigrade scale to Kelvin scale and Fahrenheit 0°C, 20°C, 120°C, 500°C, —23°C, 200°C.
Ans: Centigrade (Celsius) To Kelvin Scale:
FORMULA:

T_K = T_°C + 273

i) 0 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 0 + 273 = 273 K Ans.

ii) 20 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 20 + 273 = 293 K Ans.

iii) 120 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 120 + 273 = 393 K Ans.

iv) 500 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 500 + 273 = 773 K Ans.

v) -23 °C
Solution:
T_K = T_°C + 273
⇒ T_K = -23 + 273 = 250 K Ans.

vi) 200 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 200 + 273 = 493 K Ans.

Centigrade (Celsius) To Fahrenheit Scale:

2. Convert each of the following temperature from the Kelvin scale to the Celsius scale and Fahrenheit O K, 20 K, 100 K, 300 K, 373 K, and 500 K.
Ans: Kelvin Scale To Centigrade (Celsius):
FORMULA:

T_°C = T_K - 273

i) 0 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 0 - 273 = -273 °C Ans.

ii) 20 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 20 - 273 = -253 °C Ans.

iii) 100 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 100 - 273 = -173 °C Ans.

iv) 300 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 300 - 273 = 27 °C Ans.

v) 373 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 373 - 273 = 100 °C Ans.

vi) 500 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 500 - 273 = 227 °C Ans.

Kelvin Scale To Fahrenheit:

i) 0 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (0 - 273) + 32
⇒ T_°F = 1.8 (-273) + 32 ⇒ T_°F = - 491.4 + 32 = -459.4 °F Ans.

ii) 20 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (20 - 273) + 32
⇒ T_°F = 1.8 (-253) + 32 ⇒ T_°F = - 455.4 + 32 = -423.4 °F Ans.

iii) 100 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (100 - 273) + 32
⇒ T_°F = 1.8 (-173) + 32 ⇒ T_°F = - 311.4 + 32 = -279.4 °F Ans.

iv) 300 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (300 - 273) + 32
⇒ T_°F = 1.8 (27) + 32 ⇒ T_°F = - 48.6 + 32 = 80.6 °F Ans.

v) 373 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (373 - 273) + 32
⇒ T_°F = 1.8 (100) + 32 ⇒ T_°F = 180 + 32 = 212 °F Ans.

vi) 500 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (500 - 273) + 32
⇒ T_°F = 1.8 (227) + 32 ⇒ T_°F = 408.6 + 32 = 440.6 °F Ans.

Molecular Theory of gases - Physics II - For HSC Part 2 / XII / Class 12 (Science Group) -Short Question Answers- Section B - CRQs And Self Assessment Questions

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Unit 15: Molecular Theory of gases
Physics II -
For HSC Part 2 / XII / Class 12 (Science Group)

• Section B-CRQs
• Self Assessment Questions

SECTION B
CRQs (Constructed Response Questions)
OR Short Answered Questions)

Q.1: Why the earth is not in thermal equilibrium with the sun?
Ans: The Earth is not in thermal equilibrium with the Sun because it continuously receives energy from the Sun in the form of solar radiation. This energy heats the Earth's surface, causing temperature variations and weather phenomena. Additionally, the Earth radiates energy back into space, creating a dynamic balance rather than a static equilibrium.

OR

Ans: Earth and sun are not in equilibrium because they don't form an isolated system.

Q.2: Describe the relationship between temperature and kinetic energy of molecules.
Ans: Temperature is a measure of the average kinetic energy of the molecules in a substance. As the temperature increases, the kinetic energy of the molecules also increases, leading to faster movement and more vigorous collisions among them. Conversely, a decrease in temperature results in lower kinetic energy and slower molecular motion.

OR

Ans: The relationship between temperature and the kinetic energy of molecules is direct and proportional. As the temperature of a substance increases, the average kinetic energy of its molecules also increases. This means that at higher temperatures, molecules move faster, and at lower temperatures, they move more slowly. Temperature is essentially a measure of the average kinetic energy of the molecules in a substance.

Q.3: It is observed that when mercury in glass thermometer is put in a flame, the column of mercury first descends and then rises. Explain.
Ans: When the thermometer is first heated, the gas expands more quickly than the mercury, causing the mercury to descend initially. As the temperature continues to rise, the mercury eventually expands and rises in the tube as it absorbs heat. This phenomenon is due to the different thermal expansion properties of glass and mercury.

OR

Ans: Due to expansion of the glass, mercury first falls but later on rises due to larger coefficient of expansion for mercury than that of glass.

Q.4: What is standard temperature, pressure?
Ans: Standard temperature and pressure (STP) are defined as a temperature of 0 degra Celsius (273.15 K) and a pressure of 1 atmosphere (101.325 kPa). These conditions are used as reference points for various scientific calculations and experiments.

OR

Ans: Standard temperature and pressure (STP) refers to the nominal conditions in the atmosphere at sea level. These conditions are 0 degrees Celsius (273.15 K) and 1 atmosphere (atm) (101.325 kPa) of pressure.

Q.5: A thermometer is placed in direct sun light. What will it read the temperature?
Ans: A thermometer placed in direct sunlight may read a high temperature than the actual air temperature due to the heat absorbed from solar radiation. The reading can be influenced by factors such as the type of thermometer, its material, and exposure time.

OR

Ans: If a thermometer is placed in direct sun light, it will measure a much higher temperature than that of the air.

Q.6: The pressure in a gas cylinder containing hydrogen will leak more quickly than if it is containing oxygen. Why?
Ans: Hydrogen molecules are lighter and smaller than oxygen molecules, allowing them to move faster and escape through tiny openings more easily. This phenomenon is described by Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

OR

Ans: As the hydrogen is lighter than oxygen i.e., its molecular mass and density is less than that of oxygen therefore rate of diffusion of hydrogen gas is greater than oxygen. That is why the pressure in a gas cylinder containing hydrogen will leak more quickly than the gas cylinder containing oxygen.

Q.7: When a sealed thermos bottle full of coffee is shaken, what are the changes occur?
Ans: Shaking a sealed thermos bottle full of coffee increases the kinetic energy of the coffee molecules leading to a rise in temperature The agitation can also cause the coffee to mix more thoroughly, but since the thermos is insulated there will be minimal heat loss to the environment.

OR

Ans: When the bottle is shaken, the coffee will experience some internal movement, which may cause some molecules to collide and exchange kinetic energy. However, this effect is small, and the overall temperature of the coffee will remain nearly constant.

Q.8: How does the Kinetic theory account for the following observed facts:
(a) A gas exerts pressure
(b) The pressure of a gas depends upon its temperature.
Ans: a) A gas exerts Pressure:
According the kinetic theory, gas consists of a large number of small particles (molecules) that are in constant random motion. As these molecules collide with the walls of their container, they exert force on the walls. The cumulative effects of countless collisions results in pressure exerted by th gas. The more frequent and forceful the collisions, the higher the pressure.

b) The pressure of gas depends upon its temperature:
The kinetic theory states that the temperature of a gas is a measure of the average kinetic energy of its molecules. As the temperature increases, the kinetic energy of the molecules increases, causing them to move faster. This results in more frequent and more forceful collisions with the walls of the container, leading to an increase in pressure. Conversely lowering the temperature decreases the kinetic energy, resulting in lower pressure.

OR

Ans: The Kinetic Theory explains the behavior of gases through the motion of their particles:
(a) A gas exerts pressure:
According to the Kinetic Theory, gas particles are in constant random motion. When these particles collide with the walls of a container, they exert a force on the walls. The collective force of these collisions per unit area is what we observe as gas pressure.

(b) The pressure of a gas depends upon its temperature:
The temperature of a gas is a measure of the average kinetic energy of its particles. As the temperature increases, the particles move faster, resulting in more frequent and forceful collision with the container walls. This increase in the frequency and intensity of collisions leads to an increase i n pressure.

Q.9: Calculate the average speed of an air molecule at room temperature (20°C) and compare it to the speed of sound in air (330 m/s).
Ans: CALCULATION:
First, we need to calculate the average speed of an air molecule at room temperature.
FORMULA:
The average speed of a gas molecule can be calculated using the formula:

Where:

• K is the Kinetic energy
• k is the Boltzmann constant, approximately 1.38 x 10^-23 J/K
• T is the absolute temperature in Kelvin,
• m is the mass of a gas molecule.
  The average molar mass of air is approximately 29 g/m.
  For the mass of a single molecule. we convert this to kilograms and then divide it by Avogadro's number (6.023 x 10^-23)
  Mass of a gas molecule (m) = 29 / 1000 = 0.029 = 29 x 10^-3 Kg/ mole
  
  
• v_avg is the average speed of an air molecule
  

DATA:
T = 20 °C + 273 = 293 K
k = 1.38 x 10^-23 J/K
m = 4.82 x 10^-26 Kg
V_avg = ?

SOLUTION:

COMPARISON:
The average speed in air molecule at room temperate (20°C) is approximately 501.70 m/s, which is significantly higher than the speed of sound in air (330 m/s). This indicates that while individual molecules at high speed moving at high speeds, the collective behavior of these molecules results in sound propagation at a lower due to the interactions and compressibility of the gas.

OR

The average speed of the air molecule (501 m/s) is greater than the speed of sound in air (330 m/s)

OR

In comparison, molecule of air is moving faster than sound because V_avg = 501.70 m/s > V_sound = 330 m/s.

More Short Answered Questions

Q.10: How does Thermometer device help to measure the temperature of any other body? OR Define thermal equilibrium? OR What is the principle of thermometer?
Ans: PRINCIPLE OF THERMOMETER OR THERMAL EQUILIBRIUM:
Thermometer works on the principle of thermal equilibrium. When two bodies at different temperatures are brought in thermal contact with each other, the heat start flowing from the hot body to the cold body till the temperature of both bodies becomes same then they are said to be in thermal equilibrium.

Q.11: What do you mean by gas law
Ans: GAS LAW:
An interrelation among the physical quantities of a given sample of gas which determine the state of a gas is termed as "equation of state" of gas or Gas Laws.
These physical quantities are:

• Pressure (P)
• Volume (V)
• Temperature (T)
• Mass (m)

There are following gas laws:

• Boyle's law
• Charle's law
• Avogadro's law
• General gas equation

Q.12: What is critical point? Also define critical pressure and critical temperature?

Ans: CRITICAL POINT:
Liquid vapor equilibrium region" corresponds to the end point at the top of vaporization curve, it is called critical point and the corresponding values of P and T are called critical pressure P_C, and temperature T_C.

Critical Pressure:
Critical pressure is a liquid vapor pressure at critical temperature, where liquid and gas coexist, forming one phase..

Critical Pressure:
The critical temperature for a pure substance is the temperature above which the gas cannot become liquid, regardless of the applied pressure. A gas above the critical temperature does not separate into phases when it is compressed isothermally.

Q.13: Define Absolute zaro, Avogaro's number, real or permanent gas and ideal or perfect gas?
Ans: Absolute Zero: The temperature at which volume of a gas becomes zero and molecular motion ceases is termed as absolute zero. Kelvin selected value of this temperature as -273.15 °C on Centigrade temperature scale, −459.67 °F on the Fahrenheit temperature scale and 0K on Kelvin temperature scale

Avogadro's Number:
A mole of any substance is that mass of substance that contains a specific Number of molecules called Avogadro's number. It is represented by N_A. NA = 6.022 x 10²³ molecules/mole.
Avogadro's number is defined to be the number of carbon atoms in 12g of the isotope carbon-12. The number of moles of a substance is related to its m mass m.i.e.:
n = ^m/_M
Where M is the molecular mass of the substance usually expressed in g/mole.

Real or Permanent Gas
Real or permanent gas is a gas that obeys gas Laws at high temperature and low Pressure only.

Ideal or Perfect gas
Ideal or perfect gas is a gas that obeys gas Laws at all temperatures and pressures.

Q.14: What are K and R constant?
Ans: k Constant:
K (Boltzmann constant) is universal gas constant per molecule of a gas. Its value is 1.38 J/Molecula k.

R constant:
R is a universal gas Constant per mole of gas. Its value is 8.314 J^-1K^-1.

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SELF ASSESSMENT QUESTIONS

1. Convert each of the following temperature from the centigrade scale to Kelvin scale and Fahrenheit 0°C, 20°C, 120°C, 500°C, —23°C, 200°C.
Ans: Centigrade (Celsius) To Kelvin Scale:
FORMULA:

T_K = T_°C + 273

i) 0 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 0 + 273 = 273 K Ans.

ii) 20 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 20 + 273 = 293 K Ans.

iii) 120 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 120 + 273 = 393 K Ans.

iv) 500 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 500 + 273 = 773 K Ans.

v) -23 °C
Solution:
T_K = T_°C + 273
⇒ T_K = -23 + 273 = 250 K Ans.

vi) 200 °C
Solution:
T_K = T_°C + 273
⇒ T_K = 200 + 273 = 493 K Ans.

Centigrade (Celsius) To Fahrenheit Scale:

2. Convert each of the following temperature from the Kelvin scale to the Celsius scale and Fahrenheit O K, 20 K, 100 K, 300 K, 373 K, and 500 K.
Ans: Kelvin Scale To Centigrade (Celsius):
FORMULA:

T_°C = T_K - 273

i) 0 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 0 - 273 = -273 °C Ans.

ii) 20 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 20 - 273 = -253 °C Ans.

iii) 100 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 100 - 273 = -173 °C Ans.

iv) 300 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 300 - 273 = 27 °C Ans.

v) 373 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 373 - 273 = 100 °C Ans.

vi) 500 K
Solution:
T_°C = T_K - 273
⇒ T_°C = 500 - 273 = 227 °C Ans.

Kelvin Scale To Fahrenheit:

i) 0 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (0 - 273) + 32
⇒ T_°F = 1.8 (-273) + 32 ⇒ T_°F = - 491.4 + 32 = -459.4 °F Ans.

ii) 20 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (20 - 273) + 32
⇒ T_°F = 1.8 (-253) + 32 ⇒ T_°F = - 455.4 + 32 = -423.4 °F Ans.

iii) 100 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (100 - 273) + 32
⇒ T_°F = 1.8 (-173) + 32 ⇒ T_°F = - 311.4 + 32 = -279.4 °F Ans.

iv) 300 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (300 - 273) + 32
⇒ T_°F = 1.8 (27) + 32 ⇒ T_°F = - 48.6 + 32 = 80.6 °F Ans.

v) 373 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (373 - 273) + 32
⇒ T_°F = 1.8 (100) + 32 ⇒ T_°F = 180 + 32 = 212 °F Ans.

vi) 500 K
Solution:
T_°F = 1.8 (T_K - 273) + 32
⇒ T_°F = 1.8 (500 - 273) + 32
⇒ T_°F = 1.8 (227) + 32 ⇒ T_°F = 408.6 + 32 = 440.6 °F Ans.

3. What are the conditions (temperature and pressure) at which the triple point of water occurs?
Ans: Triple Point Of Water:
The triple point of water occurs at a temperature of 273.16 K (0.01°C) and a pressure of 4.58 mm of mercury or 611.73 Pascal.

4. Where is the triple point located in relation to the solid, liquid, and gas regions?
Ans: Location Of Triple Point Of Water:
The triple point is located at the intersection where the lines separating the solid, liquid, and gas phases meet on the phase diagram. It is the point where all three phases coexist in thermodynamic equilibrium.

5. How does the pressure of a gas change if its volume is halved at constant temperature?
Ans: According to Boyle's Law, at constant temperature, the pressure of a gas is inversely proportional to its volume. Therefore, if the volume is halved, the pressure will double.

6. What happens to the volume of a gas when its temperature is increased while keeping the pressure constant?
Ans: According to Charles's Law, at constant pressure, the volume of a gas is directly proportional to its temperature (measured in Kelvin). So, increasing the temperature will cause the volume to increase.

7. What is the relationship between the volume and the number of moles of a gas at constant temperature and pressure?
Ans: According to Avogadro's Law, at constant temperature and pressure, the volume of a gas is directly proportional to the number of moles. Increasing the number of moles will increase the volume proportionally.

8. Describe the motion of molecules in a gas according to the kinetic molecular theory.
Ans: According to the kinetic molecular theory, molecules in a gas move in all directions with a wide range of speeds. They are in constant, random motion, colliding elastically with each other and with the walls of the container. These collisions are perfectly elastic, meaning there is no loss of kinetic energy during collisions, and molecules move freely in straight lines between collisions.

9. What is the relationship between temperature and molecular motion in a gas?
The temperature of a gas is directly related to the average kinetic energy of its molecules. As temperature increases, the molecules move faster, resulting in higher average kinetic energy. Conversely, lowering the temperature decreases the molecular speed and kinetic energy.

Molecular Theory of gases - Physics II - For HSC Part 2 / XII / Class 12 (Science Group) - Section A - Multiple Choice Questions, Fill In The Blanks, Key Formulae & Concept Map

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Unit 15: Molecular Theory of gases
Physics II -
For HSC Part 2 / XII / Class 12 (Science Group)

SECTION A
Multiple Choice Questions (MCQs)

Choose the correct answer:
1. The relationship between temperature and average kinetic energy of particles in a gas is:
(a) temperature is inversely proportional to average kinetic energy
(b) temperature is directly proportional to average kinetic energy ✔
(c) temperature is independent of average kinetic energy
(d) temperature is proportional to the square of average kinetic energy

2. Standard conditions of temperature and pressure (STP) refer to a gas at:
(a) 0 °C and 1 atm ✔
(b) 20 °C and 1 atm
(c) 25 °C and 1 atm
(d) 30 °C and 101.3 kPa (1 atm)

3. If the temperature is kept constant and the volume of a gas is doubled, then pressure of a gas is:
(a) Reduced to ¹/₂ of the original value. ✔
(b) Doubled
(c) Reduced to ¹/₄ of the original value
(d) Quadrupled

4. The Avogadro's number is the number of molecules in:
(a) One mole of a substance ✔
(b) One kg of a substance
(c) One m³ of a gas
(d) One kilogram of hydrogen gas

5. Mean translational K.E. per molecule of an ideal gas at temperature T is:
(a)  ³/₂ kT ✔
(b) ¹/₂ kT
(c) ²/₃ kT
(d) kT⁴

6. The normal human body temperature is:
(a) 98.6 °F (37°C) ✔
(b) 99.6 °F (37.4 °C)
(c) 100.4 °F (38 °C)
(d) 101 °F (38.3 °C)

7. The pressure P, the density 𝝆 and the average speed of molecules of an ideal gas are related by the equation.
(a) P = ²/₃ ml²
(b) P = ¹/₃ ml²
(c) P = ¹/₃ 𝝆 V² ✔
(d) P = ²/₃ l²

8. In air at S.T.P, the average speed of the:
(a) Oxygen molecules is greater than Nitrogen molecules
(b) Nitrogen molecules is greater than Oxygen molecules
(c) Oxygen molecules is approximately equal to Nitrogen molecules
(d) Helium atoms is greater than both Oxygen and Nitrogen molecules ✔

9. If the absolute temperature of a gas is increased 3 times, the rms velocity of the molecules will be:
(a) 3 times
(b) 9 times
(c) √3 times ✔
(d) ¹/₃ times

10. A gas is enclosed in an isolated container which is placed on a fast-moving train uniformly. The temperature of the gas:
(a) Increases due to the motion of the train
(b) Decreases due to the motion of the train
(c) Remains constant ✔
(d) Fluctuates, depending on the train's speed and direction

More Multiple Choice Questions (MCQs)

TEMPERATURE
11. -459.4°F is the temperature at which. volume and pressure of the real gas becomes:
a) maximum
b) constant
c) minimum
d) zero ✔

12.The average kinetic energy of the molecules of the object is called its:
a) specific heat
b) temperature ✔
c) heat
d) hotness

13. 4.180 J of work raises the temperature of one litre of water through:
a) 1 K
b) 1 °C ✔
c) 1 °F
d) 2 °C

14. The thermal state of a body is defined by:
a) Temperature ✔
b) Heat
c) Cold
d) none of these

15. S.I. unit of temperature is:
a) Fahrenheit
b) Kelvin ✔
c) Celsius
d) Joule

16. Freezing point of water is:
a) 273 K
b) 0 °C
c) 32 °F
d) all of these ✔

17. The Fahrenheit and Kelvin scales will have the same reading when the temperature is:
a) 474.25°
b) 374.25°
c) 574.25° ✔
d) 274.25°

18. Hotness or cold of an object is expressed in terms of:
a) Heat
b) Temperature ✔
c) Thermal energy
d) Chemical energy

19. The tempeture of body shows its:
a) Physical state
b) Degree of hotness
c) Degree of coldness
d) All of these ✔

20. If heat energy is removed from an object, its temperature will normally:
a) Not change
b) Rise
c) Fall ✔
d) Fall, then rise

21. The relation between centigrade and, Fahrenheit scales of temperature is:
a) T°C = 9/5(T°F — 32)
b) T°C = 5/9(T°F + 32)
c) T°C = 5/9 (T°F — 32) ✔
d) T°F = 5/9 (T°C — 32)

22. The temperature of a normal human body is 98.6 °F. This temperature on centigrade is:
a) 0 °C
b) 37 °C ✔
c) 57 °C
d) 30 °C

23. This one of the following is the correct:
a) 1 °F < 1°C ✔
b) 1°F > 1°C
c) 1 °F = 1 °C
d) none of these

24. This one of the following is correct:
a) 1°C < IK
b) 1°C > IK
c) 1°C = IK ✔
d) none of these

25. The temperature of human body on Kelvin scale is:
a) 273 K
b) 373 K
c) 310 K ✔
d) 236 K

26. Absolute zero corresponds to this temperature on Fahrenheit scale:
a) 32 °F
b) —180 °F
c) —460 °F ✔
d) 212 °F

27. On Fahrenheit scale, the temperature of 50°C will be:
a) 40 °F
b) 10 °F
c) 122 °F ✔
d) 105 °F

28. Fahrenheit and Celsius scales of temptiqure coincide at:
a) 0°
b) 273°
c) —273°
d) —40° ✔

29. Two ends "A" & "B" of a rod are at temperature -10 °C and —30°C. The heat will flow:
a) from —30°C to —10°C
b) not at all
c) from 10°C to —30°C ✔
d) none of these

30. 273 K is equal to:
a) 0 °F
b) —32 °F
c) —273 °F
d) 32 °F ✔

31. On Celsius scale, 1°C in magnitude is equal to:
a) 32 °F
b) 16 °F
c) 0 °F
d) 1.8 °F ✔

32. The absolute temperature, corresponding to 212°F, is:
a) 485 K
b) 373 K ✔
c) 161 K
d) 100 K

KINETIC MOLECULAR THEORY OF GASES
33. The K.E. of the molecules of an ideal gas at absolute zero will be:
a) very low
b) very high
c) zero ✔
d) below zero

34. Kinetic energy of gas molecule is:
a) RT
b) ³/₂ RT
c) KT
d) ³/₂ KT ✔

35. Kinetic energy per mole of ideal gas is:
a) RT
b) ³/₂ RT ✔
c) KT
d) ³/₂ KT

36. Upon this physical quantity, Kinetic energy of gas molecules depends:
a) Temperature of gas ✔
b) Pressure of gas
c) Volume of gas
d) Nos. of moles of gas

37. The universal gas constant per molecule is called:
a) Raleigh-Jean's constant
b) Boltzman's constant ✔
c) Gas constant
d) Stefan's constant

38. At volume V and temperature T, the pressure of Nmolecules of the gas is P. If the number of molecules of the gas is doubled at constant volume, then:
a) Temperatute will become half
b) Temperature will become twice
(c) Pressure will become half
d) Pressure will become twice ✔

39. The root me speed of the molecules of an ideal gas in a sealed container is V. The gas is heated until the pressure in the container is trebled. The r.m.s. speed is now:
a) √3 V ✔
b) V/9
c) 9V
d) 3V

40. The average K.E. of the molecules of an ideal in a closed vessel is increased by a factor 4. The pressure of the gas:
a) will remain the same
b) will increase by factor of 2
c) will increase by a factor of 4 ✔
d) will increase by a factor of √2

41. A fixed mass of a gas is heated at constant volume, this one of the following will not change:
a) Temperature
b) Molecular motion
c) Internal energy
d) Density of the gas ✔

42. Pressure of an ideal gas can be written in terms of its diensity 𝝆:
a) P = 𝝆v²
b) P = ²/₃ 𝝆v²
c) P = ¹/₃ 𝝆v² ✔
b) P = ¹/₂ 𝝆v²

43. At constant temperature, the pressure P of an ideal gas is proportional to its density 𝝆 as:
a) P ∝ ¹/ _𝝆
b) P ∝ 𝝆 ✔
c) P ∝ ¹/ _𝝆²
d) P ∝ 𝝆²

44. The average translational kinetic energy of the molecules of a gas at absolute is proportional to:
a) ¹/ _T
b) T ✔
c) √T
d) T²

45. The unit of the pressure of gas is:
a) 1N/m²
b) One Pascal
c) One atmosphere
d) all of these ✔

46. According to the Kinetic theory of gases, the absolute temperature of a perfect gas is:
(a) Inversely proportional to the K.E. of the molecules
b) Independent of kinetic energy of the molecules
c) Equal to the kinetic energy of the molecules
d) Directly proportional to the average translational kinetic energy of the molecules ✔

47. The average internal energy of an ideal gas is called:
a) Pressure
b) Volume
c) Temperature ✔
d) Heat

48. A gas is enclosed in a cylinder of volume "V" pressure "P" and absolute temperature "T". If a mass of each molecule is "m" then density of gas is:
a) mK/T
b) mR/T
c) mP/KT ✔
d) mp/RT

BOYLE'S LAW, CHARLE'S LAW AND GENERAL GAS EQUATION
49. For a gas, obeying Boyle's law, if the pressure is doubled, the volume becomes:
a) Double
b) One half ✔
c) Four times
d) One fourth

50. Boyles law holds for ideal gases in:
a) isochoric process
b) isobaric process
c) isothermal process ✔
d) adiabatic process

51. If the volume, of a gas is held constiat and we increase its temperatur then:
a) Its pressure is constant
b) Its pressure rises ✔
c) Its pressure falls
d) None of these

52. The pressure exerted by a column of mercury 76cm high and 0 °C is called:
a) 1 atmosphere ✔
b) 1 Newton per square meter
c) 1 Pascal
d) Data is insufficient

53. Standard condition STP refers to a gas at:
a) 76 cm Hg 0°C
b)1 atm 273K
c) 760 mm of Hg 273K
d) All of these ✔

54. The relation between the pressure P and Volume V of fixed mass of gas at constant temperature is:
a) P/V = Constant
b) PV² constant
c) PV = Constant ✔
d) P²V = Constant

55. According to Boyle's law, the volume of a given mass of a gas is:
a) Directly proportional to the temperature
b) Directly proportional to the pressure
c) Inversely proportional to the mass
d) Inversely proportional to the Pressure at constant temperature ✔

56. The graph Boyle's. law is a:
a) Curve
b) Slope
c) Straight line
d) Hyperbola ✔

57. If the volume of a given mass of a gas is doubled wiithout changing its temperature, the pressure of the gas is:
a) Reduce to ¹/₂ of the initial value ✔
b) The same as the initial value
c) Reduced to ¹/₄ of the initial value
d) Double of the initial value

58. This one of the following curves represents Boyle's law:
59. Charles' Law states that the volume of' a given mass of gas is:
a) Inversely proportional to the absolute temperature at constant pressure
b) Directly proportional to the absolute temperature at constant pressure ✔
c) Direcily proportional to the pressure
d) Inversely proportional to the pressure

60. Charles' Law can b written mathematically as:
a) V ∝ T ✔
b) V ∝ 1/T
c) P ∝ V
d) P ∝ 1/V

61. The volume of a given gas at constant pressure becomes zero at:
a) 273 K
b) 273 °C
c) 373 K
d) -273 °C ✔

62 The. graph of Charles' law is a:
a) Curve
b) Slope
c) Straight line ✔
d) Hyperbola

63. A single equation that relates all the three variables of gas is given by:
a) PT/V = constant
b) PV/T = constant ✔
c) VT/P = constant
d) none of these

64. The ideal gas law is given in the form:
a) PV = nR /T
b) PT = NRV
c) PV = nRT ✔
d) TV = nRP

65. The value of universal gas constant `R' in S.1. unit is:
a) 8.310 J/mol K
b) 8314 J/mol K
c) 831.4 J/mol K
d) 8.314 J/mol K ✔

66. A gas which strictly obeys the laws under all conditions of temperature and pressure is:
a) Real gas
b) Ideal gas ✔
c) Permanent gas
d) Insert gas

67. At constant pressure,The volume of the given mass of a gas is V at temperature T. At this temperature, the volume of the gas will he 4V:
a) 4T ✔
b) ¹/₄ T
c) 2T
d) ¹/₂ T

68. A container is filled with a sample of an ideal gas at a pressure of 1.5 atm. The gas is compressed isothermally to one-fourth of its original volume. Its new pressure will be:
a) 2 atm
b) 4 atm
c) 6 atm ✔
d) 9 atm

Fill In The Blanks

1. Our study of Mechanics is based on the three fundamental quantities mass, length and time.
2. Temperature is a measure of the average kinetic energy of particles in a substance.
3. A temperature scale that starts at absolute zero is the theoretical lowest possible temperature. It is measured in Kelvin (K).
4. The triple point of water is the temperature (0.01°C) and pressure (611.73 Pascal's) at which water coexists in all three phases (solid, liquid, and gas).
5. If two systems are each in thermal equilibrium (equal temperature) with a third system, then the two initial systems are in thermal equilibrium with each other.
6. The general gas law states that PV = nRT, relating pressure, volume, number of moles, and temperature of a gas.
7. Boyle's Law: P₁V₁ = P₂V₂
8. According to Boyle's law Pressure and volume are inversely proportional at constant temperature.
9. Charle's Law: V₁ / T₁ = V₂ / T₂
10. According to Charle's Law Volume and temperature are directly proportional at constant pressure.
11. General Gas Law: V = nRT / P
12. According to general gas law Volume and number of moles are directly proportional at constant temperature and pressure.
13. KTG stands for Kinetic Theory of Gases.
14. Gases consist of tiny particles (molecules) that are in constant motion.
15. The molecules are very small compared to the distance between them.
16. According to Kinetic Theory Gases the molecules are in constant random motion.
17. The molecules collide with each other and the container walls.
18. Molecular movement and pressure by the equation: P = (2/3) nkT.
19. The pressure exerted by a gas is caused by the collisions of molecules with the container walls, and is related to the temperature and volume of the gas.
20. The kinetic energy of molecules is directly proportional to the temperature of the gas.
21. The kinetic energy is described by the equation KE = (3/2) kT.
22. 1 dm = 0.1m
23. 1 dm³ = 0.1m

Key Formulae

Special Thanks To Sir Wasim Ahmad
ICS Virtual Digital Coaching Center

Concept Map

Definitions To Remember

• Volume: The space occupied by a substance is called its Volume.
• Pressure: The force exerted by gas molecules on walls of vessels due to collision is called pressure.