Question 1

An ideal gas is enclosed in a thermally insulated sealed container. During an adiabatic expansion, the mean interval between successive molecular collisions varies with the volume V as V^q. What is the value of q in terms of γ = Cp/Cv?

Accepted Answer

(γ + 1)/2. Mean free path lambda ~ V. Adiabatic: T ~ V^(1-gamma), so v ~ sqrt(T) ~ V^((1-gamma)/2). Collision interval ~ lambda/v ~ V^(1-(1-gamma)/2) = V^((gamma+1)/2), so q = (gamma+1)/2.

Question 2

At 300 K, the root mean square speed of hydrogen molecules is 1930 m/s. What will be the r.m.s. speed of oxygen molecules at 1200 K?

Accepted Answer

965 m/s. v_rms scales as sqrt(T/M). Ratio = sqrt((1200/300)*(2/32)) = sqrt(4 * 1/16) = sqrt(0.25) = 0.5. So v_O2 = 1930 * 0.5 = 965 m/s.

Question 3

A gaseous mixture contains 2 moles of oxygen and 4 moles of argon at temperature T. If vibrational degrees of freedom are ignored, what is the total internal energy of the mixture?

Accepted Answer

11 RT. The total internal energy of an ideal gas is calculated using the formula U = (f/2) nRT, where f is the degrees of freedom. For this mixture, oxygen has 5 degrees of freedom (3 translational + 2 rotational) and argon has 3 degrees of freedom (3 translational), leading to a total of 11 degrees of freedom for 6 moles of gas, resulting in an internal energy of 11 RT.

Question 4

If all intermolecular attractions were removed, what volume would be occupied by the molecules present in 4.5 g of water at standard temperature and pressure?

Accepted Answer

5.6 litre. If intermolecular attractions vanish the molecules behave as an ideal gas at STP. 4.5 g water = 0.25 mol, occupying 0.25*22.4 = 5.6 litre.

Question 5

An ideal gas sample has volume V, pressure P, and absolute temperature T. If the mass of one molecule is m, which expression gives the density of the gas?

Accepted Answer

Pm/kT. From PV = NkT, number of molecules N = PV/(kT). Mass = N*m = PVm/(kT), so density = mass/V = Pm/(kT).

Question 6

A vessel at temperature T contains N molecules of gas A, each having mass m, and 2N molecules of gas B, each having mass 2m. If the mean square speed of molecules of gas B is V2 and that of gas A is V1, then the ratio V1/V2 equals

Accepted Answer

2. V1 = _A = 3kT/m and V2 = _B = 3kT/(2m). So V1/V2 = (3kT/m)/(3kT/2m) = 2.

Question 7

The speed of sound in oxygen (O₂) at a certain temperature is 460 m s⁻¹. The speed of sound in helium (He) at the same temperature will be (assume both gases to be ideal)

Accepted Answer

1421 m s⁻¹. The speed of sound in a gas is inversely related to the square root of its molar mass; helium, being much lighter than oxygen, allows sound to travel faster through it, resulting in a significantly higher speed of sound.

Question 8

One kg of a diatomic gas is at a pressure of 8 × 10⁴ N/m². The density of the gas is 4 kg/m³. What is the energy of the gas due to its thermal motion?

Accepted Answer

5 × 10⁴ J. The energy of a diatomic gas due to its thermal motion can be calculated using the formula for internal energy, which is related to the number of degrees of freedom and the temperature. Given the pressure and density, the correct calculation leads to an energy of 5 × 10⁴ J, confirming option A as the right answer.

Question 9

Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as √q, where V is the volume of the gas. The value of q is
(γ = Cₚ/C_v)

Accepted Answer

(γ+1)/2. The correct option is derived from the relationship between the average time of collision and the volume of the gas during adiabatic expansion, where the factor q is related to the specific heat ratio γ. In this context, the expression (γ+1)/2 accurately represents how the average time of collision scales with the volume of the gas.

Question 10

The mass of a hydrogen molecule is 3.32×10⁻²⁷ kg. If 10²³ hydrogen molecules strike, per second, a fixed wall of area 2 cm² at an angle of 45° to the normal, and rebound elastically with a speed of 10³ m/s, then the pressure on the wall is nearly:

Accepted Answer

2.35 × 10³ N/m². Only the normal momentum component reverses: per molecule dp = 2mv cos45. Force = N*2mv cos45 = 1e23 * 2 * 3.32e-27 * 1e3 * 0.707 = 0.4695 N. Pressure = F/A = 0.4695 / 2e-4 = 2.35e3 N/m^2.

Question 11

A mixture of 2 moles of helium gas (atomic mass = 4u), and 1 mole of argon gas (atomic mass = 40u) is kept at 300 K in a container. The ratio of their rms speeds [V_rms (helium) / V_rms (argon)] is close to:

Accepted Answer

3.16. The root mean square speed of a gas is inversely proportional to the square root of its molar mass. Since helium has a much lower molar mass than argon, its rms speed is significantly higher, resulting in a ratio of approximately 3.16.

Question 12

For a given gas at 1 atm pressure, rms speed of the molecules is 200 m/s at 127°C. At 2 atm pressure and at 227°C, the rms speed of the molecules will be:

Accepted Answer

100√5 m/s. The root mean square (rms) speed of gas molecules is proportional to the square root of the temperature divided by the pressure. By applying the formula for rms speed and considering the changes in both temperature and pressure, we find that at 2 atm and 227°C, the rms speed calculates to 100√5 m/s.

Question 13

An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is v̄, m is its mass and k_B is Boltzmann constant, then its temperature will be:

Accepted Answer

mv̄²/3k_B. The rms speed is set by translational motion: (1/2) m vbar^2 = (3/2) k_B T, so T = m vbar^2/(3 k_B). Rotational and vibrational modes do not enter the rms velocity.

Question 14

The temperature of an open room of volume 30 m³ increases from 17°C to 27°C due to the sunshine. The atmospheric pressure in the room remains 1 × 10⁵ Pa. In n_i and n_f are the number of molecules in the room before and after heating, then n_f - n_i =

Accepted Answer

-2.5 × 10²⁵. The correct option is -2.5 × 10²⁵ because as the temperature increases, the air expands, causing a decrease in the number of molecules in the room if the volume remains constant and the pressure is held steady. This results in a negative change in the number of molecules, indicating a loss.

Question 15

Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is: (Atomic wt. of Cl = 35.5 u) [JEE-Main On line-2018]

Accepted Answer

0.46. At same T and P, density is proportional to molar mass. Ratio = M(NH3)/M(HCl) = 17/36.5 = 0.46.

Question 16

The value closed to the thermal velocity of a Helium atom at room temperature (300K) in ms⁻¹ is -
[kB = 1.4 × 10⁻²³ J/K; mHe = 7 × 10⁻²⁷ kg]

[JEE-Main On line-2018]

Accepted Answer

1.3 × 10³. The thermal velocity of a gas particle can be calculated using the formula v = sqrt(3kT/m), where k is the Boltzmann constant, T is the temperature, and m is the mass of the particle. Substituting the values for helium at room temperature yields a result close to 1.3 × 10³ ms⁻¹, making this the correct option.

Question 17

A 15 g mass of nitrogen gas is enclosed in a vessel at a temperature 27°C. A amount of heat transferred to the gas, so that rms velocity of molecules is doubled, is about: [Take R = 8.3 J/K mole]

Accepted Answer

10 kJ. n = 15/28 mol. Doubling v_rms quadruples T (v ~ sqrt(T)), so dT = 4*300 - 300 = 900 K. For diatomic N2, Cv = (5/2)R, giving Q = (15/28)(2.5*8.3)(900) ~ 10 kJ.

Question 18

For a given at 1 atm pressure, rms speed of the molecules is 200 m/s at 127°C. At 2 atm pressure and at 227°C, the rms speed of the molecules will be-

Accepted Answer

100√5 m/s. The root mean square (rms) speed of gas molecules is proportional to the square root of the temperature divided by the molar mass and inversely proportional to the square root of the pressure. By applying the ideal gas law and the relationship between temperature and pressure, we can determine that at 2 atm and 227°C, the rms speed is 100√5 m/s.

Question 19

An HCl molecule has rotational, translational and vibrational motions. If the rms velocity of HCl molecules in its gaseous phase is v̄, m is its mass and kB is Boltzmann constant, then its temperature will be -

Accepted Answer

m v̄² / 3kB. The correct option is based on the equipartition theorem, which states that each degree of freedom contributes to the total energy. For a diatomic molecule like HCl, there are three translational and two rotational degrees of freedom, leading to a total of five degrees of freedom. However, the effective temperature relation for gases considers only the translational motion, which results in the factor of 3 in the denominator when relating rms velocity to temperature.

Question 20

The specific heats Cₚ and C_v of a gas of diatomic molecules, A, are given (in units of J mol⁻¹ K⁻¹) by 29 and 22, respectively. Another gas of diatomic molecules, B, has the corresponding values 30 and 21. If they are treated as ideal gases, then:

Accepted Answer

A has a vibrational mode but B has none.. Cv = (f/2)R. Gas A has Cv = 22 -> f ~ 5.3, above the rigid-diatomic value 5, so A has a vibrational mode. Gas B has Cv = 21 ~ 5R/2 = 20.8 -> f ~ 5, so B is rigid with no vibrational mode. Hence A has a vibrational mode but B has none.

Question 21

An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature 300 K. The mean time between two successive collisions is 6 × 10⁻⁸ s. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to

Accepted Answer

4 × 10⁻⁸ s. Mean free time tau = lambda/v ~ (T/P)/sqrt(T) = sqrt(T)/P. tau2 = tau1 * sqrt(T2/T1) * (P1/P2) = 6e-8 * sqrt(500/300) * (1/2) = 3.87e-8 s, closest to 4e-8 s.

Question 22

A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is:

Accepted Answer

15. The total internal energy of an ideal gas mixture can be calculated using the formula U = (f/2) * nRT, where f is the degrees of freedom and n is the total number of moles. For oxygen, which has a rigid bond, the degrees of freedom is 5 (3 translational and 2 rotational), and for argon, it is 3 (only translational). Thus, the total internal energy for the mixture is calculated as (5/2 * 3 + 3/2 * 5)RT = 15RT.

Question 23

An ideal gas in a closed container is slowly heated. As its temperature increases, which of the following statements are true ?
(A) the mean free path of the molecules decreases.
(B) the mean collision time between the molecules decreases.
(C) the mean free path remains unchanged.
(D) the

Accepted Answer

(B) and (C). Volume and N are fixed, so number density and hence mean free path are unchanged (C). Higher T raises mean speed, so mean collision time = lambda/v_avg decreases (B). Answer: (B) and (C).

Question 24

Match the Cₚ/C_v ratio for ideal gases with different type of molecules:

Accepted Answer

(4) (A)-(IV), (B)-(I), (C)-(II), (D)-(III). gamma values: monoatomic = 5/3 (IV), diatomic rigid = 7/5 (I), diatomic non-rigid (with vibration, f=7) = 9/7 (II), triatomic rigid (f=6) = 4/3 (III). So (A)-IV, (B)-I, (C)-II, (D)-III.

Question 25

Number of molecules in a volume of 4 cm³ of a perfect monoatomic gas at some temperature T and at a pressure of 2 cm of mercury is close to ? (Given, mean kinetic energy of a molecule (at T) is 4 × 10⁻¹⁴ erg, g = 980 cm/s², density of mercury = 13.6 g/cm³)

Accepted Answer

4.0 × 10¹⁸. P = rho*g*h = 13.6*980*2 = 26656 dyne/cm^2; PV = 26656*4 = 1.066e5 erg. kT = (2/3)(4e-14) = 2.67e-14 erg. N = PV/kT = 1.066e5/2.67e-14 ~ 4.0e18.

Question 26

Molecules of an ideal gas are known to have three translational degrees of freedom and two rotational degrees of freedom. The gas is maintained at a temperature of T. The total internal energy, U of a mole of this gas, and the value of γ ( = Cₚ/C_v) are given respectively, by

Accepted Answer

U = 5/2 RT and γ = 7/5. The internal energy of an ideal gas is determined by its degrees of freedom, which for a diatomic gas includes three translational and two rotational degrees, leading to U = (5/2)RT. The ratio of specific heats, γ, is derived from the relationship between heat capacities, and for a diatomic gas, it is 7/5.

Question 27

In a dilute gas at pressure P and temperature T, the mean time between successive collisions of a molecule varies with T as:

Accepted Answer

√T. At constant pressure, number density n = P/(kT) ~ 1/T, so the mean free path lambda ~ 1/n ~ T. Mean speed ~ sqrt(T), hence mean time between collisions tau = lambda/v ~ T/sqrt(T) = sqrt(T).

Question 28

Consider two identical diatomic gases A and B at same temperature T. Molecules of the gas A are rigid, and have a mass m. Molecules of the gas B have an additional vibrational mode, and have a mass m/4. The ratio of the specific heats (C_V^A and C_V^B) of gas A and B, respectively, is-

Accepted Answer

5: 7. Gas A (rigid diatomic) has Cv=5/2 R. Gas B (diatomic + one vibrational mode) has Cv=5/2 R + R = 7/2 R. The ratio Cv_A:Cv_B = (5/2):(7/2) = 5:7.

Question 29

Calculate the value of mean free path (λ) for oxygen molecules at temperature 27°C and pressure 1.01 × 10⁵ Pa. Assume the gas molecular diameter is 0.3 nm and the gas is ideal. (k = 1.38 × 10⁻²³ J K⁻¹)

Accepted Answer

102 nm. The mean free path (λ) is calculated using the formula λ = kT / (√2πd²P), where k is the Boltzmann constant, T is the temperature in Kelvin, d is the molecular diameter, and P is the pressure. Substituting the given values results in a mean free path of 102 nm, confirming that option D is correct.

Question 30

If one mole of the polyatomic gas is having two vibrational modes and β is the ratio of molar specific heats for polyatomic gas [β = Cp/Cv], then the value of β is:

Accepted Answer

1.2. Degrees of freedom f = 3 (trans) + 3 (rot) + 2x2 (two vibrational modes) = 10. So Cv = (f/2)R = 5R, Cp = 6R, and beta = Cp/Cv = 6/5 = 1.2.

Question 31

What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature T? (kB is Boltzmann constant)

Accepted Answer

1/2 kB T. In an ideal gas, each degree of freedom contributes an average energy of 1/2 kB T due to the equipartition theorem, which states that energy is equally distributed among all degrees of freedom.

Question 32

Consider a sample of oxygen behaving like an ideal gas. At 300 K, the ratio of root mean square (rms) velocity to the average velocity of gas molecule would be: (Molecular weight of oxygen is 32 g/mol; R = 8.3 J K⁻¹ mol⁻¹)

Accepted Answer

√(3π/8). vrms = sqrt(3RT/M) and vavg = sqrt(8RT/(pi*M)). Their ratio is sqrt(3/(8/pi)) = sqrt(3*pi/8), independent of T and M.

Question 33

The correct relation between the degrees of freedom f and the ratio of specific heat γ is:

Accepted Answer

(1) f = 2/(γ − 1). The correct relation f = 2/(γ − 1) arises from the kinetic theory of gases, where the degrees of freedom f are related to the specific heat ratio γ, reflecting how energy is distributed among the translational and internal modes of motion in a gas.

Question 34

Consider a mixture of gas molecule of types A, B and C having masses mA < mB < mC. The ratio of their root mean square speeds at normal temperature and pressure is:

Accepted Answer

1/vA < 1/vB < 1/vC. Root-mean-square speed is proportional to 1/sqrt(m). Since mAvB>vC, which means 1/vA<1/vB<1/vC.

Question 35

What will be the average value of energy for a monoatomic gas in thermal equilibrium at temperature T ?

Accepted Answer

3/2 k_B T. The average energy of a monoatomic gas in thermal equilibrium is derived from the equipartition theorem, which states that each degree of freedom contributes 1/2 k_B T to the energy. A monoatomic gas has three translational degrees of freedom, leading to a total average energy of 3/2 k_B T.

Question 36

On the basis of kinetic theory of gases, the gas exerts pressure because its molecules:

Accepted Answer

suffer change in momentum when impinging on the walls of container.. The correct option is right because, according to the kinetic theory of gases, gas molecules collide with the walls of the container, and these collisions result in a change in momentum, which is what generates pressure.

Question 37

The root mean square speed of molecules of a given mass of a gas at 27°C and 1 atmosphere pressure is 200 m s⁻¹. The root mean square speed of molecules of the gas at 127°C and 2 atmosphere pressure is x/√3 m s⁻¹. The value of x will be _____.

Accepted Answer

400. rms speed depends only on temperature: v2/v1 = sqrt(T2/T1) = sqrt(400/300) = 2/sqrt3. So v2 = 200 x 2/sqrt3 = 400/sqrt3 m/s. Thus x = 400 (pressure has no effect).

Question 38

Given below are two statements:
Statement I: In a diatomic molecule, the rotational energy at a given temperature obeys Maxwell's distribution.
Statement II: In a diatomic molecule, the rotational energy at a given temperature equals the translational kinetic energy for each molecule.
In t

Accepted Answer

Both Statement I and Statement II are false.. Rotational energy follows a Boltzmann (not Maxwell) distribution; Maxwell's distribution describes molecular speeds, so Statement I is false. Rotational energy does not equal translational KE per molecule (translational uses 3 degrees of freedom = (3/2)kT, rotational of a diatomic uses 2 = kT, equal only on average per degree of freedom), so Statement II is false.

Question 39

The rms speeds of the molecules of Hydrogen, Oxygen and Carbondioxide at the same temperature are V_H, V_O and V_C respectively then:

Accepted Answer

V_H > V_O > V_C. The root mean square (rms) speed of gas molecules is inversely proportional to the square root of their molar mass. Since hydrogen has the lowest molar mass among the three gases, it has the highest rms speed, followed by oxygen and then carbon dioxide.

Question 40

If the rms speed of oxygen molecules at 0°C is 160 m/s, find the rms speed of hydrogen molecules at 0°C.

Accepted Answer

640 m/s. At equal temperature v_rms scales as 1/sqrt(M). With M(O2)=32 and M(H2)=2, v_H = v_O*sqrt(32/2) = 160*4 = 640 m/s.

Question 41

A mixture of hydrogen and oxygen has volume 500 cm³, temperature 300 K, pressure 400 kPa and mass 0.76 g. The ratio of masses of oxygen to hydrogen will be:-

Accepted Answer

16: 3. The correct option is based on the molar masses of hydrogen and oxygen, where the molar mass of hydrogen (H2) is approximately 2 g/mol and that of oxygen (O2) is about 32 g/mol. By applying the ideal gas law and the given conditions, we can derive the mass ratio of oxygen to hydrogen, which simplifies to 16:3.

Question 42

0.056 kg of Nitrogen is enclosed in a vessel at temperature of 127°C. The amount of heat required to double the speed of its molecules is __ k cal. (Take R = 2 cal mol⁻¹ K⁻¹)

Accepted Answer

12.00. To double the speed of the nitrogen molecules, we need to increase the temperature by a factor of four, since the speed of gas molecules is proportional to the square root of temperature. The heat required can be calculated using the formula Q = nC_vΔT, where n is the number of moles, C_v is the molar heat capacity at constant volume, and ΔT is the change in temperature, leading to the conclusion that 12.00 kcal is needed.

Question 43

Following statements are given:
(A) The average kinetic energy of a gas molecule decreases when the temperature is reduced.
(B) The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature.
(C) The average kinetic energy of a gas molecule decreas

Accepted Answer

(A) and (D) only. Average kinetic energy of a gas molecule depends only on temperature: (A) is true (lower T -> lower KE) and (D) is true (P rises with T at constant volume). (B), (C) and (E) are false since KE is independent of P and V. Correct: (A) and (D) only.

Question 44

Sound travels in a mixture of two moles of helium and n moles of hydrogen. If rms speed of gas molecules in the mixture is √2 times the speed of sound, then the value of n will be

Accepted Answer

2. gamma_mix = (2*Cp_He + n*Cp_H2)/(2*Cv_He + n*Cv_H2) = (5+3.5n)/(3+2.5n) = 1.5. Solving gives 0.5 = 0.25n, so n = 2.

Question 45

A flask contains argon and oxygen in the ratio of 3: 2 in mass and the mixture is kept at 27°C. The ratio of their average kinetic energy per molecule respectively will be

Accepted Answer

1: 1. The average kinetic energy per molecule of a gas is directly proportional to its temperature and is independent of the type of gas. Since both argon and oxygen are at the same temperature, their average kinetic energies per molecule are equal, resulting in a ratio of 1:1.

Question 46

According to kinetic theory of gases,
A. The motion of the gas molecules freezes at 0°C
B. The mean free path of gas molecules decreases if the density of molecules is increased.
C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.
D. Aver

Accepted Answer

B and C only. Option B is correct because increasing the density of gas molecules leads to more frequent collisions, thus reducing the mean free path. Option C is also correct as increasing the temperature at constant pressure results in higher kinetic energy and greater mean free path due to increased molecular speed.

Question 47

A vessel contains 14 g of nitrogen gas at a temperature of 27°C. The amount of heat to be transferred to the gas to double its r.m.s. speed of its molecules will be; Take R = 8.32 J mol⁻¹ K⁻¹.

Accepted Answer

9360 J. 14 g N2 = 0.5 mol at 300 K. Doubling rms speed makes T = 4×300 = 1200 K, so ΔT = 900 K. With diatomic Cv = 5R/2, Q = 0.5 × (5/2)(8.32) × 900 = 9360 J.

Question 48

The root mean square speed of smoke particles of mass 5 × 10⁻¹⁷ kg in their Brownian motion in air at NTP is approximately. [Given k = 1.38 × 10⁻²³ JK⁻¹]

Accepted Answer

15 mm s⁻¹. The root mean square speed of particles in Brownian motion can be calculated using the formula v_rms = sqrt(3kT/m), where k is the Boltzmann constant, T is the temperature in Kelvin, and m is the mass of the particles. At normal temperature and pressure (NTP), the calculated speed for the given mass of smoke particles yields approximately 15 mm/s, confirming option C as the correct answer.

Question 49

Match List I with List II:

List I
(A) 3 Translational degrees of freedom
(B) 3 Translational, 2 rotational degrees of freedoms
(C) 3 Translational, 2 rotational and 1 vibrational degrees of freedom
(D) 3 Translational, 3 rotational and more than one vibrational degrees of freedom

List II

Accepted Answer

(A)-(I), (B)-(III), (C)-(IV), (D)-(II). The correct option accurately matches the degrees of freedom of different types of gases: monoatomic gases have only translational motion, rigid diatomic gases have translational and rotational motion, nonrigid diatomic gases add vibrational motion, and polyatomic gases have even more vibrational modes.

Question 50

Three vessels of equal volume contain gases at the same temperature and pressure. The first vessel contains neon (monoatomic), the second contains chlorine (diatomic) and third contains uranium hexafluoride (polyatomic). Arrange these on the basis of their root mean square speed (v_rms) an

Accepted Answer

v_rms (mono) > v_rms (dia) > v_rms (poly). The root mean square speed of a gas is inversely proportional to the square root of its molar mass. Since neon is monoatomic and has a lower molar mass compared to chlorine (diatomic) and uranium hexafluoride (polyatomic), it will have the highest v_rms, followed by chlorine, and then uranium hexafluoride, which has the highest molar mass.

JEE Main Physics: Kinetic Theory questions with solutions

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