NEET Physics: Heat, Thermodynamics and Kinetic Theory · Chapter 3

Kinetic Theory

What to remember

Pressure of an ideal gas: P = ⅓ρv²rms = ⅓(N/V)m v²rms. The average kinetic energy of a molecule is (3/2)k_BT, which depends only on the absolute temperature.
Speeds are in the order v_mp < v_avg < v_rms, with v_rms = √(3k_BT/m) = √(3RT/M), v_avg = √(8k_BT/πm) and v_mp = √(2k_BT/m).
Law of equipartition: each quadratic term in the energy of a molecule has an average energy of ½k_BT. For f degrees of freedom, Cv = (f/2)R, Cp = (f/2 + 1)R and γ = 1 + 2/f. This gives γ = 5/3 (monatomic), 7/5 (rigid diatomic) and 4/3 (rigid nonlinear polyatomic).
Molar specific heat of a solid is 3R (Dulong-Petit); the mean free path of a gas is l = 1/(√2 π n d²).

Molecular nature of matter

All matter is made of atoms and molecules. In a solid they are close together and vibrate about fixed positions. In a liquid they are close but can move past one another. In a gas they are far apart, move freely and collide with one another and with the walls. The size of a gas molecule is about 1 Å (10⁻¹⁰ m), and in a gas at ordinary conditions the average distance between molecules is about ten times larger. The number of particles in one mole is the Avogadro number, N_A = 6.022 × 10²³ mol⁻¹.

Behaviour of gases

Law	Statement	Condition
Boyle's law	PV = constant	T and amount of gas constant
Charles' law	V/T = constant	P and amount of gas constant
Avogadro's law	Equal volumes of all gases contain equal numbers of molecules	same T and P
Dalton's law	Total pressure is the sum of the partial pressures of the component gases	gases that do not react

Combining these gives the ideal-gas equation PV = nRT, with R = 8.314 J mol⁻¹ K⁻¹. If N is the number of molecules, PV = Nk_BT, where k_B = 1.38 × 10⁻²³ J K⁻¹ is the Boltzmann constant and R = N_A k_B. At standard temperature and pressure (0 °C and 1 atm) one mole of any ideal gas occupies about 22.4 L. A real gas behaves like an ideal gas at low pressure and high temperature.

Kinetic theory of an ideal gas

Assumptions:

1. A gas contains a very large number of identical molecules in random motion.
2. The size of a molecule is very small compared with the average distance between molecules.
3. Molecules exert no force on each other except during collisions, so there is no potential energy between them.
4. Collisions between molecules and with the walls are perfectly elastic.
5. The time of a collision is very small compared with the time between collisions.
6. The motion is random, so all directions are equally likely.

Pressure. The pressure is caused by the repeated collisions of molecules with the wall. For N molecules of mass m in volume V:

P = ⅓(N/V) m v²rms = ⅓ρ v²rms

Here v²rms is the mean of the squares of the speeds, and ρ is the density of the gas. Pressure can also be written as P = ⅔ × (kinetic energy per unit volume).

How the pressure formula arises. A molecule of mass m moving with speed v_x towards a wall and rebounding elastically changes its momentum by 2mv_x. The number of such collisions per second on a wall of area A is proportional to the number density and to v_x. Adding the contributions of all molecules gives a pressure proportional to n m v_x². Because the motion is random, the averages of v_x², v_y² and v_z² are equal, so each is one-third of v²rms. This is the source of the factor ⅓. The pressure is larger when there are more molecules per unit volume, when they are heavier, or when they are faster.

Kinetic interpretation of temperature. Using PV = Nk_BT, the average translational kinetic energy per molecule is

½ m v²rms = (3/2) k_BT

The temperature is a measure of the average kinetic energy of the molecules and does not depend on the type of gas. At the same temperature, molecules of different gases have the same average translational kinetic energy. At 0 K this motion would stop. For one mole of a monatomic ideal gas, the internal energy is U = (3/2)RT.

Speeds of gas molecules

Speed	Formula	Relative value
Most probable, v_mp	√(2k_BT/m)	1 (smallest)
Average, v_avg	√(8k_BT/πm)	about 1.13
Root mean square, v_rms	√(3k_BT/m) = √(3RT/M)	about 1.22 (largest)

So v_mp : v_avg : v_rms = √2 : √(8/π) : √3. All three are proportional to √T and inversely proportional to √m (or √M, where M is the molar mass). At the same temperature, hydrogen molecules are much faster than oxygen molecules.

Law of equipartition of energy and degrees of freedom

Law of equipartition: in thermal equilibrium at absolute temperature T, every independent quadratic term in the expression for the energy of a molecule has an average value of ½k_BT. The number of such independent ways is the number of degrees of freedom f.

Type of molecule	Translational	Rotational	Vibrational (each mode counts 2)	Total f
Monatomic (He, Ne, Ar)	3	0	0	3
Diatomic, rigid (O₂, N₂ at ordinary temperature)	3	2	0	5
Diatomic, with vibration	3	2	2	7
Polyatomic, nonlinear, rigid	3	3	0	6

A vibrating mode counts for 2 because it has both a kinetic and a potential energy term, each of ½k_BT. The average energy per molecule is (f/2)k_BT, and the internal energy of one mole is (f/2)RT.

Specific heat capacities of gases

For one mole of an ideal gas, U = (f/2)RT, so:

Cv = (f/2)R
Cp = Cv + R = (f/2 + 1)R
γ = Cp/Cv = 1 + 2/f

Gas	f	Cv	Cp	γ
Monatomic	3	3R/2	5R/2	5/3 (about 1.67)
Diatomic, rigid	5	5R/2	7R/2	7/5 (1.40)
Diatomic, with vibration	7	7R/2	9R/2	9/7 (about 1.29)
Polyatomic, nonlinear, rigid	6	3R	4R	4/3 (about 1.33)

The more degrees of freedom, the smaller the value of γ. The values for monatomic gases agree well with experiment. For diatomic gases at ordinary temperatures, the experimental values agree with the rigid model because vibration is not active at these temperatures.

Specific heat capacity of solids and water

Solids. An atom in a solid vibrates in three directions about its mean position. Each direction has one kinetic and one potential term, so there are 6 quadratic terms. The average energy per atom is 3k_BT, and the internal energy of one mole is U = 3RT. Hence the molar specific heat is

C = 3R ≈ 24.9 J mol⁻¹ K⁻¹

This is the Dulong-Petit law. It holds for most solids at ordinary temperatures but fails at low temperatures. The specific heat per unit mass is c = 3R/M, where M is the molar mass in kg.

Water. Treating the three atoms of a water molecule as a solid-like system gives 9R ≈ 74.8 J mol⁻¹ K⁻¹. The measured value is about 75.4 J mol⁻¹ K⁻¹, which is close.

Mean free path

The mean free path l is the average distance a molecule travels between two successive collisions. For molecules of diameter d and number density n (molecules per unit volume):

l = 1/(√2 π n d²)

Using P = nk_BT, this becomes l = k_BT/(√2 π d² P). Therefore:

l increases when the temperature increases at constant pressure.
l decreases when the pressure (or number density) increases at constant temperature.
l decreases when the molecular diameter increases.

In air at ordinary conditions, l is of the order of 10⁻⁷ m, far larger than the molecular diameter.

Worked examples

Example 1 (rms speed). Find the rms speed of oxygen at 27 °C. M = 32 g mol⁻¹ = 0.032 kg mol⁻¹, R = 8.3 J mol⁻¹ K⁻¹. v_rms = √(3RT/M) = √(3 × 8.3 × 300/0.032) = √(2.33 × 10⁵) ≈ 483 m s⁻¹.

Example 2 (kinetic energy). At 300 K the average translational kinetic energy per molecule is (3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 300 = 6.2 × 10⁻²¹ J.

Example 3 (pressure). A gas of density 1.2 kg m⁻³ has v_rms = 500 m s⁻¹. P = ⅓ρv²rms = ⅓ × 1.2 × 500² = 1.0 × 10⁵ Pa.

Example 4 (temperature and speed). The rms speed of a gas at 27 °C (300 K) is v. At what temperature will it be 2v? Since v_rms ∝ √T, the temperature must be 4 times: 4 × 300 = 1200 K, which is 927 °C.

Example 5 (comparing gases). At the same temperature, v_rms ∝ 1/√M. For hydrogen (M = 2 g mol⁻¹) and oxygen (M = 32 g mol⁻¹), v_H/v_O = √(32/2) = 4. The two gases have the same average kinetic energy per molecule, but the hydrogen molecules move four times faster.

Example 6 (heat at constant volume). Find the heat needed to raise the temperature of 2 mol of a rigid diatomic gas by 10 K at constant volume (R = 8.3 J mol⁻¹ K⁻¹). Cv = 5R/2, so Q = nCvΔT = 2 × 2.5 × 8.3 × 10 = 415 J.

Example 7 (Dulong-Petit). For aluminium, M = 27 g mol⁻¹. c = 3R/M = 24.9/0.027 ≈ 920 J kg⁻¹ K⁻¹, which is close to the measured value of about 900.

Formula summary

Quantity	Formula
Ideal gas	PV = nRT = Nk_BT
Pressure	P = ⅓ρv²rms
Mean kinetic energy per molecule	(3/2)k_BT
rms speed	√(3k_BT/m) = √(3RT/M)
Average speed	√(8k_BT/πm)
Most probable speed	√(2k_BT/m)
Energy per mole	U = (f/2)RT
Specific heats	Cv = (f/2)R, Cp = (f/2 + 1)R, γ = 1 + 2/f
Solid (molar)	C = 3R
Mean free path	l = 1/(√2 π n d²)

NEET traps

1. v_rms, v_avg and v_mp are different. v_rms is the largest and v_mp the smallest.
2. Mean kinetic energy per molecule depends only on T, not on the type of gas; speed depends on both T and mass.
3. A rigid diatomic molecule has f = 5, not 7; the vibrational modes count only when stated.
4. A vibrational mode contributes 2 degrees of freedom, not 1.
5. γ becomes smaller as f becomes larger.
6. Do not forget the factor √2 in the mean free path.
7. The mean free path increases with temperature at constant pressure but decreases with pressure at constant temperature.
8. v_rms is proportional to √T, so doubling T raises the speed by √2, not 2.
9. Pressure is P = ⅓ρv²rms; do not use ½ and do not omit the square.
10. The Dulong-Petit value 3R is a molar value; the specific heat per kilogram depends on the molar mass.

One-liners

1. The ideal-gas equation is PV = nRT = Nk_BT.
2. k_B = 1.38 × 10⁻²³ J K⁻¹.
3. N_A = 6.022 × 10²³ mol⁻¹.
4. One mole of an ideal gas at STP occupies about 22.4 L.
5. P = ⅓ρv²rms.
6. Mean kinetic energy per molecule = (3/2)k_BT.
7. v_rms ∝ √(T/M).
8. v_mp : v_avg : v_rms = √2 : √(8/π) : √3.
9. Monatomic gas: f = 3 and γ = 5/3.
10. Rigid diatomic gas: f = 5 and γ = 7/5.
11. Rigid nonlinear polyatomic gas: f = 6 and γ = 4/3.
12. Each quadratic energy term carries ½k_BT.
13. Molar specific heat of a solid is 3R.
14. Molar specific heat of water is about 9R.
15. Mean free path l = 1/(√2 π n d²).

Practice questions

The average translational kinetic energy of a molecule of an ideal gas at absolute temperature T is
1. (3/2)k_BT
2. (5/2)k_BT
3. (1/2)k_BT
4. 3k_BT
Answer
A. (3/2)k_BT
By the kinetic theory ½mv²rms = (3/2)k_BT for each molecule, whatever the gas.
The ratio Cp/Cv for a monatomic ideal gas is
1. 4/3
2. 7/5
3. 5/3
4. 9/7
Answer
C. 5/3
f = 3 gives γ = 1 + 2/3 = 5/3.
Take k_B = 1.38 × 10⁻²³ J K⁻¹. The average translational kinetic energy of a molecule of an ideal gas at 300 K is about
1. 6.2 × 10⁻²³ J
2. 6.2 × 10⁻²¹ J
3. 1.2 × 10⁻²⁰ J
4. 4.1 × 10⁻²¹ J
Answer
B. 6.2 × 10⁻²¹ J
(3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 300 = 6.2 × 10⁻²¹ J. (k_BT alone is 4.1 × 10⁻²¹ J.)
At a given temperature, the correct order of the speeds of molecules of an ideal gas is
1. v_avg < v_mp < v_rms
2. v_rms < v_avg < v_mp
3. v_mp < v_rms < v_avg
4. v_mp < v_avg < v_rms
Answer
D. v_mp < v_avg < v_rms
The speeds are in the ratio √2 : √(8/π) : √3, or about 1 : 1.13 : 1.22.
A gas of density 1.2 kg m⁻³ has an rms molecular speed of 500 m s⁻¹. Its pressure is
1. 1.5 × 10⁵ Pa
2. 3.0 × 10⁵ Pa
3. 3.3 × 10⁴ Pa
4. 1.0 × 10⁵ Pa
Answer
D. 1.0 × 10⁵ Pa
P = ⅓ρv²rms = ⅓ × 1.2 × 250 000 = 1.0 × 10⁵ Pa. (½ in place of ⅓ would give 1.5 × 10⁵ Pa.)
The number of degrees of freedom of a rigid diatomic molecule is
1. 3
2. 6
3. 7
4. 5
Answer
D. 5
A rigid diatomic molecule has 3 translational and 2 rotational degrees of freedom.
The molar specific heat at constant volume of a monatomic ideal gas is
1. 3R
2. 3R/2
3. 5R/2
4. R/2
Answer
B. 3R/2
Cv = (f/2)R with f = 3, so Cv = 3R/2.
For a gas with γ = 4/3, the number of degrees of freedom per molecule is
1. 3
2. 6
3. 5
4. 7
Answer
B. 6
γ = 1 + 2/f gives 2/f = 1/3, so f = 6.
At the same temperature, the ratio of the rms speed of hydrogen molecules (M = 2 g mol⁻¹) to that of oxygen molecules (M = 32 g mol⁻¹) is
1. 16
2. 2
3. 1/4
4. 4
Answer
D. 4
v_rms ∝ 1/√M, so the ratio = √(32/2) = √16 = 4.
The rms speed of a gas at 27 °C is v. At what temperature does the rms speed become 2v?
1. 927 °C
2. 54 °C
3. 327 °C
4. 600 °C
Answer
A. 927 °C
v_rms ∝ √T. For 2v, T becomes 4 × 300 = 1200 K, which is 927 °C.
The molar specific heat at constant pressure of a rigid diatomic gas is
1. 3R
2. 9R/2
3. 7R/2
4. 5R/2
Answer
C. 7R/2
Cp = Cv + R = 5R/2 + R = 7R/2.
The pressure of an ideal gas is doubled at constant temperature. The mean free path of its molecules becomes
1. half
2. double
3. four times
4. unchanged
Answer
A. half
l ∝ 1/n ∝ 1/P at fixed T. Doubling P halves l.
The volume of one mole of an ideal gas at STP is 22.4 L and N_A = 6.022 × 10²³ mol⁻¹. The number of molecules in 1 L at STP is about
1. 6.0 × 10²³
2. 2.7 × 10²⁵
3. 2.7 × 10²²
4. 2.7 × 10¹⁹
Answer
C. 2.7 × 10²²
N = 6.022 × 10²³/22.4 = 2.7 × 10²². (2.7 × 10²⁵ is the number per m³.)
Avogadro's law states that equal volumes of all gases at the same temperature and pressure contain
1. equal masses
2. equal numbers of molecules
3. equal numbers of atoms
4. equal densities
Answer
B. equal numbers of molecules
This is the statement of Avogadro's law.
Take R = 8.3 J mol⁻¹ K⁻¹. The internal energy of 2 mol of a monatomic ideal gas at 300 K is
1. 4980 J
2. 12 450 J
3. 7470 J
4. 2490 J
Answer
C. 7470 J
U = (3/2)nRT = 1.5 × 2 × 8.3 × 300 = 7470 J.
Given below are two statements: Statement I: The average translational kinetic energy of the molecules of an ideal gas depends only on the absolute temperature. Statement II: At the same temperature, molecules of different gases have the same average translational kinetic energy.
1. Both Statement I and Statement II are incorrect
2. Both Statement I and Statement II are correct
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are correct
Both follow from (3/2)k_BT, which has no dependence on the type of gas.
Given below are two statements: Statement I: Among the rms, average and most probable speeds, the rms speed is the greatest. Statement II: v_rms = √(3k_BT/m).
1. Both Statement I and Statement II are incorrect
2. Both Statement I and Statement II are correct
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are correct
v_mp : v_avg : v_rms = √2 : √(8/π) : √3. Both statements are correct.
Given below are two statements: Statement I: The pressure of an ideal gas is P = ⅓ρv²rms. Statement II: The pressure of a gas is caused by the weight of the gas molecules resting on the walls.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is incorrect but Statement II is correct
4. Statement I is correct but Statement II is incorrect
Answer
D. Statement I is correct but Statement II is incorrect
Pressure comes from momentum transferred in collisions with the walls, not from weight.
Given below are two statements: Statement I: The number of degrees of freedom of a rigid diatomic gas molecule is 5. Statement II: A monatomic gas molecule has 3 rotational degrees of freedom.
1. Both Statement I and Statement II are correct
2. Statement I is correct but Statement II is incorrect
3. Both Statement I and Statement II are incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Statement I is correct but Statement II is incorrect
A monatomic molecule has only 3 translational degrees of freedom.
Which of the following statements are correct? 1. Each quadratic energy term carries an average energy of ½k_BT per molecule. 2. A vibrational mode of a diatomic molecule counts as two degrees of freedom. 3. A rigid nonlinear polyatomic gas has γ = 5/3.
1. 1 and 3 only
2. 2 and 3 only
3. 1 and 2 only
4. 1, 2 and 3
Answer
C. 1 and 2 only
For a rigid nonlinear polyatomic gas f = 6 and γ = 4/3, so 3 is wrong.
Which of the following statements are correct? 1. The mean free path increases when the temperature increases at constant pressure. 2. The mean free path decreases when the molecular diameter increases. 3. The mean free path is independent of pressure at constant temperature.
1. 2 and 3 only
2. 1 and 3 only
3. 1, 2 and 3
4. 1 and 2 only
Answer
D. 1 and 2 only
l = k_BT/(√2 π d² P). It rises with T, falls with d², and falls with P, so 3 is wrong.
Given below are two statements: Statement I: The molar specific heat of most solids at ordinary temperature is about 3R. Statement II: Each atom in a solid has six quadratic energy terms (three kinetic and three potential).
1. Both Statement I and Statement II are incorrect
2. Both Statement I and Statement II are correct
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Both Statement I and Statement II are correct
The atom oscillates in three directions, giving 6 terms and an energy of 3k_BT per atom, so C = 3R.
Given below are two statements: Statement I: At the same temperature, lighter gas molecules have a greater rms speed than heavier ones. Statement II: At the same temperature, they have the same average translational kinetic energy.
1. Both Statement I and Statement II are correct
2. Both Statement I and Statement II are incorrect
3. Statement I is correct but Statement II is incorrect
4. Statement I is incorrect but Statement II is correct
Answer
A. Both Statement I and Statement II are correct
v_rms ∝ 1/√m, while the mean kinetic energy (3/2)k_BT is the same for all gases.
Given below are two statements: Statement I: Boyle's law states that V is proportional to T at constant pressure. Statement II: Charles' law states that V is proportional to 1/P at constant temperature.
1. Both Statement I and Statement II are correct
2. Statement I is correct but Statement II is incorrect
3. Both Statement I and Statement II are incorrect
4. Statement I is incorrect but Statement II is correct
Answer
C. Both Statement I and Statement II are incorrect
These are swapped. Boyle: PV constant at constant T. Charles: V/T constant at constant P.
Given below are two statements: Statement I: In the kinetic theory, collisions of molecules with the walls are taken as perfectly elastic. Statement II: In the kinetic theory, molecules exert large attractive forces on each other even when they are far apart.
1. Both Statement I and Statement II are correct
2. Statement I is correct but Statement II is incorrect
3. Both Statement I and Statement II are incorrect
4. Statement I is incorrect but Statement II is correct
Answer
B. Statement I is correct but Statement II is incorrect
The theory assumes that molecules exert no force on each other except during collisions.
Assertion (A): The pressure of a gas in a closed rigid vessel rises when it is heated. Reason (R): The average kinetic energy of the molecules increases, so they hit the walls harder and more often.
1. Both A and R are true but R is not the correct explanation of A
2. A is true but R is false
3. Both A and R are true and R is the correct explanation of A
4. A is false but R is true
Answer
C. Both A and R are true and R is the correct explanation of A
At constant V, P ∝ T. A higher speed means more momentum per collision and more collisions. R explains A.
Assertion (A): Hydrogen molecules have a greater rms speed than oxygen molecules at the same temperature. Reason (R): The rms speed is inversely proportional to the square root of the molar mass.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
v_rms = √(3RT/M), and M of hydrogen is smaller. R explains A.
Assertion (A): The temperature of a gas is a measure of the average kinetic energy of its molecules. Reason (R): The molecules of a gas move randomly.
1. Both A and R are true and R is the correct explanation of A
2. A is true but R is false
3. A is false but R is true
4. Both A and R are true but R is not the correct explanation of A
Answer
D. Both A and R are true but R is not the correct explanation of A
A is true as (3/2)k_BT is the mean energy. R is true but does not explain the link with temperature.
Assertion (A): The mean free path of a gas decreases when it is compressed at constant temperature. Reason (R): The mean free path is directly proportional to the number density of molecules.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
C. A is true but R is false
A is true. R is false: l = 1/(√2 π n d²), so l is inversely proportional to n.
Assertion (A): The average velocity (a vector) of the molecules of a gas at rest in a closed vessel is not zero. Reason (R): The molecules move in all directions with equal probability.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is false but R is true
4. A is true but R is false
Answer
C. A is false but R is true
A is false: the average velocity is zero. R is true, and it is the reason the average velocity is zero.
Assertion (A): The value of γ for a rigid diatomic gas is smaller than that for a monatomic gas. Reason (R): A diatomic molecule has more degrees of freedom than a monatomic molecule, and γ = 1 + 2/f.
1. Both A and R are true and R is the correct explanation of A
2. Both A and R are true but R is not the correct explanation of A
3. A is true but R is false
4. A is false but R is true
Answer
A. Both A and R are true and R is the correct explanation of A
f = 5 gives 7/5 and f = 3 gives 5/3. More degrees of freedom give a smaller γ. R explains A.
Match List I with List II. List I: a. Root mean square speed b. Most probable speed c. Average speed d. Mean kinetic energy per molecule List II: i. √(2k_BT/m) ii. √(8k_BT/πm) iii. √(3k_BT/m) iv. (3/2)k_BT
1. a-iii, b-i, c-ii, d-iv
2. a-iii, b-i, c-iv, d-ii
3. a-i, b-iii, c-ii, d-iv
4. a-iii, b-ii, c-i, d-iv
Answer
A. a-iii, b-i, c-ii, d-iv
v_rms: iii. v_mp: i. v_avg: ii. Mean kinetic energy: iv.
Match List I with List II. List I: a. Monatomic gas b. Rigid diatomic gas c. Diatomic gas with vibration d. Rigid nonlinear polyatomic gas List II: i. 7 ii. 3 iii. 6 iv. 5
1. a-iv, b-ii, c-i, d-iii
2. a-ii, b-iv, c-iii, d-i
3. a-ii, b-i, c-iv, d-iii
4. a-ii, b-iv, c-i, d-iii
Answer
D. a-ii, b-iv, c-i, d-iii
Degrees of freedom: monatomic 3 (ii), rigid diatomic 5 (iv), with vibration 7 (i), rigid polyatomic 6 (iii).
Match List I with List II. List I: a. Root mean square speed depends on b. Mean kinetic energy per molecule depends on c. Mean free path depends on d. Molar specific heat of solids List II: i. 3R for all simple solids ii. Number density and molecular diameter iii. Temperature and molar mass iv. Temperature only
1. a-iii, b-iv, c-ii, d-i
2. a-iii, b-ii, c-iv, d-i
3. a-iv, b-iii, c-ii, d-i
4. a-iii, b-iv, c-i, d-ii
Answer
A. a-iii, b-iv, c-ii, d-i
v_rms: T and M (iii). Mean KE: T only (iv). l: n and d (ii). Solids: 3R (i).
Match List I with List II. List I: a. Boltzmann constant b. Avogadro number c. Molar volume at STP d. Gas constant List II: i. 6.022 × 10²³ mol⁻¹ ii. 22.4 L iii. 1.38 × 10⁻²³ J K⁻¹ iv. 8.314 J mol⁻¹ K⁻¹
1. a-i, b-iii, c-ii, d-iv
2. a-iii, b-ii, c-i, d-iv
3. a-iii, b-i, c-iv, d-ii
4. a-iii, b-i, c-ii, d-iv
Answer
D. a-iii, b-i, c-ii, d-iv
k_B: iii. N_A: i. Molar volume: ii. R: iv.
Match List I with List II. List I: a. Boyle's law b. Charles' law c. Avogadro's law d. Dalton's law List II: i. V/T constant at constant P ii. PV constant at constant T iii. Equal volumes contain equal numbers of molecules iv. Total pressure is the sum of partial pressures
1. a-ii, b-i, c-iii, d-iv
2. a-i, b-ii, c-iii, d-iv
3. a-ii, b-iii, c-i, d-iv
4. a-ii, b-i, c-iv, d-iii
Answer
A. a-ii, b-i, c-iii, d-iv
Boyle: ii. Charles: i. Avogadro: iii. Dalton: iv.
Match List I with List II. List I: a. Cv of a monatomic gas b. Cp of a monatomic gas c. Cp of a rigid diatomic gas d. Cp of a rigid nonlinear polyatomic gas List II: i. 7R/2 ii. 3R/2 iii. 4R iv. 5R/2
1. a-ii, b-iv, c-iii, d-i
2. a-ii, b-iv, c-i, d-iii
3. a-ii, b-i, c-iv, d-iii
4. a-iv, b-ii, c-i, d-iii
Answer
B. a-ii, b-iv, c-i, d-iii
3R/2 (ii); 5R/2 (iv); 7R/2 (i); 4R (iii).
Take R = 8.3 J mol⁻¹ K⁻¹. The rms speed of oxygen molecules (M = 32 g mol⁻¹) at 27 °C is about
1. 240 m s⁻¹
2. 960 m s⁻¹
3. 15 m s⁻¹
4. 480 m s⁻¹
Answer
D. 480 m s⁻¹
v = √(3RT/M) = √(3 × 8.3 × 300/0.032) ≈ 483 m s⁻¹. Using M in grams instead of kilograms gives 15 m s⁻¹, which is wrong.
The rms speed of oxygen (M = 32 g mol⁻¹) at 320 K is the same as the rms speed of hydrogen (M = 2 g mol⁻¹) at a temperature of
1. 20 K
2. 5120 K
3. 80 K
4. 1280 K
Answer
A. 20 K
v_rms ∝ √(T/M), so T/M is the same: T = 320 × 2/32 = 20 K.
Take R = 8.3 J mol⁻¹ K⁻¹. Using the Dulong-Petit law, the specific heat capacity of aluminium (M = 27 g mol⁻¹) is about
1. 307 J kg⁻¹ K⁻¹
2. 24.9 J kg⁻¹ K⁻¹
3. 920 J kg⁻¹ K⁻¹
4. 2490 J kg⁻¹ K⁻¹
Answer
C. 920 J kg⁻¹ K⁻¹
c = 3R/M = 24.9/0.027 ≈ 920 J kg⁻¹ K⁻¹.
A gas has a number density of 2.5 × 10²⁵ m⁻³ and a molecular diameter of 3 × 10⁻¹⁰ m. Take π = 3.14 and √2 = 1.414. The mean free path is about
1. 1.4 × 10⁻⁷ m
2. 1.0 × 10⁻⁷ m
3. 1.0 × 10⁻⁵ m
4. 1.0 × 10⁻⁹ m
Answer
B. 1.0 × 10⁻⁷ m
l = 1/(√2 π n d²) = 1/(1.414 × 3.14 × 2.5 × 10²⁵ × 9 × 10⁻²⁰) = 1.0 × 10⁻⁷ m. Omitting √2 gives 1.4 × 10⁻⁷ m.
Take R = 8.3 J mol⁻¹ K⁻¹. The heat needed to raise the temperature of 2 mol of a rigid diatomic gas by 10 K at constant volume is
1. 415 J
2. 581 J
3. 830 J
4. 249 J
Answer
A. 415 J
Q = nCvΔT = 2 × (5/2) × 8.3 × 10 = 415 J.
A mixture contains equal numbers of moles of a monatomic gas and a rigid diatomic gas. The ratio Cp/Cv of the mixture is
1. 23/15
2. 6/5
3. 7/5
4. 3/2
Answer
D. 3/2
Cv = (3R/2 + 5R/2)/2 = 2R and Cp = 3R, so γ = 3/2. The mean of the two γ values (23/15) is wrong.
Take k_B = 1.38 × 10⁻²³ J K⁻¹. At a pressure of 1.0 × 10⁵ Pa and a temperature of 300 K, the number of molecules per cubic metre of an ideal gas is about
1. 2.4 × 10²⁸
2. 7.2 × 10²⁷
3. 2.4 × 10²⁵
4. 2.4 × 10²³
Answer
C. 2.4 × 10²⁵
n = P/(k_BT) = 10⁵/(1.38 × 10⁻²³ × 300) = 2.4 × 10²⁵ m⁻³.

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1.The caloric theory replaced which older theory of heat and combustion?
2.The sum of the kinetic energy and the potential energy of an object is called ______.
3."Theory of Natural selection" was proposed by -
4.The theory behind stars twinkling is that
5.The kinetic energy E possessed by an object of mass m and moving with a uniform velocity v is _____.
6.The kinetic energy of an object of mass m moving with a velocity v is 25 J. What will be its kinetic energy when its velocity is increased three times (3v)?
7.The kinetic energy possessed by a body of mass 2 kg is 100 J. The velocity of the body will be:
8.On increasing the temperature of solids, the kinetic energy of the particles ________.
9.Kinetic energy will become/remain________ when the momentum of the body is doubled.
10.Kinetic energy of gas molecules is directly proportional to-

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