Kinetic Theory of Gases and ldeal Gas | Physics JEE previous year questions with Solutions

Q 21 :
The work done in an adiabatic change in an ideal gas depends upon only: [2025]

change in its pressure
change in its specific heat
change in its volume
change in its temperature

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(4)

Work done in adiabatic process,

$W = \frac{n R △ T}{1 - γ}$

Q 22 :
Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.

Reason (R) : In isothermal process, PV = constant, while in adiabatic process $P V^{γ}$ = constant. Here $γ$ is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas.

In the light of the above statements, choose the correct answer from the options given below: [2025]

Both (A) and (R) are true but (R) is NOT the correct explanation of (A)
(A) is true but (R) is false
Both (A) and (R) are true and (R) is the correct explanation of (A)
(A) is false but (R) is true

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(4)

For isothermal process

PV = constant

$\Rightarrow P d V + V d P = 0$

$\Rightarrow \frac{d V}{V} = \frac{- d P}{P}$

$\Rightarrow d V = - (\frac{V}{P}) d P$

For adiabatic $P V^{γ}$ = constant

$\Rightarrow P_{γ} V^{γ - 1} d V + V^{γ} d P = 0$

$\Rightarrow d V = \frac{- V}{γ^{P}} (d P)$

Magnitude of $| \frac{V}{P} |$ is greater than $| \frac{V}{γ^{P}} |$

So, volume falls more rapidly in adiabatic.

2nd statement is the process description of the isothermal and adiabatic.

Q 23 :

A poly-atomic molecule ( $C_{V}$ = 3R, $C_{P}$ = 4R, where R is gas constant) goes from phase space point A ( $P_{A} = 10^{5} Pa, V_{A} = 4 \times 10^{- 6} m^{3}$ ) to point B ( $P_{B} = 5 \times 10^{4} Pa, V_{B} = 6 \times 10^{- 6} m^{3}$ ) to point C ( $P_{C} = 10^{4} Pa, V_{C} = 8 \times 10^{- 6} m^{3}$ ). A to B is an adiabatic path and B to C is an isothermal path.

The net heat absorbed per unit mole by the system is:. [2025]

500 R (ln 3 + ln 4)
450 R (ln 4 – ln 3)
500 R ln 2
400 R ln 4

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(2)

For process A $\to$ B

${(△ Q)}_{A B} = 0$ (adiabatic)

For process B $\to$ C

${(△ Q)}_{B C} = {(△ W)}_{B C} = n R T \ln (\frac{V_{C}}{V_{B}})$

= 450 R [ln (4) – ln 3]

${(△ Q)}_{net} = {(△ Q)}_{A B} + {(△ Q)}_{B C}$

= 450 R [ln (4) – ln (3)]

Q 24 :
Identify the characteristics of an adiabatic process in a monoatomic gas.

(A) Internal energy is constant.

(B) Work done in the process is equal to the change in internal energy.

(C) The product of temperature and volume is a constant.

(D) The product of pressure and volume is a constant.

(E) The work done to change the temperature from $T_{1}$ to $T_{2}$ is proportional to ( $T_{2} - T_{1}$ )

Choose the correct answer from the options given below: [2025]

(A), (C), (D) only
(A), (C), (E) only
(B), (E) only
(B), (D) only

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(3)

For adiabatic process, Q = 0

$Q = △ U + W = 0 \Rightarrow W = - △ U$

$W = - n C_{v} △ T \Rightarrow | W | = n C_{v} △ T \propto T_{2} - T_{1}$

B, E are correct

Q 25 :
During the melting of a slab of ice at 273 K at atmospheric pressure: [2025]

Internal energy of ice-water system remains unchanged.
Positive work is done by the ice-water system on the atmosphere.
Internal energy of the ice-water system decreases.
Positive work is done on the ice-water system by the atmosphere.

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(4)

Melting of ice requires heat therefore internal energy increases. But because of decrease in volume the work done on atmosphere is negative or atmosphere does positive work on ice-water system.

Q 26 :
A piston of mass M is hung from a massless spring whose restoring force law goes as $F = - k x^{3}$ , where k is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with 'n' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height $L_{0}$ to $L_{1}$ , the total energy delivered by the filament is (Assume spring to be in its natural length before heating) [2025]

$3 n R T \ln (\frac{L_{1}}{L_{0}}) + 2 M g (L_{1} - L_{0}) + \frac{k}{3} (L_{1}^{3} - L_{0}^{3})$
$n R T \ln (\frac{L_{1}^{2}}{L_{0}^{2}}) + \frac{M g}{2} (L_{1} - L_{0}) + \frac{k}{4} (L_{1}^{4} - L_{0}^{4})$
$n R T \ln (\frac{L_{1}}{L_{0}}) + M g (L_{1} - L_{0}) + \frac{k}{4} (L_{1}^{4} - L_{0}^{4})$
$n R T \ln (\frac{L_{1}}{L_{0}}) + M g (L_{1} - L_{0}) + \frac{3 k}{4} (L_{1}^{4} - L_{0}^{4})$

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(3)

Total energy suplied = gravitational potential energy of piston + spring potential energy + work done by gas.

Q $= M g (h_{2} - h_{1}) + \int_{L_{0}}^{L_{1}} k x^{3} d x + μ R T \ln \frac{V_{2}}{V_{1}}$

$= M g (L_{1} - L_{0}) + \frac{k}{4} (L_{1}^{4} - L_{0}^{4}) + n R T \ln [\frac{L_{1} A}{L_{0} A}]$

$= M g (L_{1} - L_{0}) + \frac{k}{4} (L_{1}^{4} - L_{0}^{4}) + n R T \ln [\frac{L_{1}}{L_{0}}]$

Q 27 :
A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800 ${cm}^{3}$ and temperature 27°C. The change in temperature when the gas is adiabatically compressed to 200 ${cm}^{3}$ is:

(Take $γ$ = 1.5 : $γ$ is the ratio of specific heats at constant pressure and at constant volume)

327 K
600 K
522 K
300 K

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(4)

$V_{1} = 800 {cm}^{3}, V_{2} = 200 {cm}^{3}, T_{1} = 300 K$

for adiabatic process, $T V^{γ - 1}$ = const.

$(300) {(800)}^{1.5 - 1} = T_{2} {(200)}^{1.5 - 1}$

$T_{2} = 300 {[\frac{800}{200}]}^{0.5} = 300 \times {(2^{2})}^{1 / 2}$

$T_{2} = 600 K$

$△ T = 600 - 300 = 300 K$

Q 28 :
An ideal gas exists in a state with pressure $P_{0}$ , volume $V_{0}$ . It is isothermally expanded to 4 times of its initial volume ( $V_{0}$ ), then isobarically compressed to its original volume. Finally the system is heated isochorically to bring it to its initial state. The amount of heat exchanged in this process is: [2025]

$P_{0} V_{0} (2 \ln 2 - 0.75)$
$P_{0} V_{0} (\ln 2 - 0.75)$
$P_{0} V_{0} (\ln 2 - 0.25)$
$P_{0} V_{0} (2 \ln 2 - 0.25)$

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(1)

$Q = △ U + W$

For cyclic process Q = W

$W = W_{1} + W_{2} + W_{3}$

$= P_{0} V_{0} \ln \frac{4 V_{0}}{V_{0}} + \frac{P_{0}}{4} (V_{0} - 4 V_{0}) + 0$

$= P_{0} V_{0} \ln 4 - \frac{3}{4} P_{0} V_{0}$

$= P_{0} V_{0} (2 \ln 2 - 0.75)$

Q 29 :
Match List I wuth List II

List List

(A) Isobaric (I) $△ Q = △ W$

(B) Isochoric (II) $△ Q = △ U$

(C) Adiabatic (III) $△ Q = zero$

(D) Isothermal (IV) $△ Q = △ U + P △ V$

$△ Q$ = Heat supplied

$△ W$ = Work done by the system

$△ U$ = Change in internal energy

P = Pressure of the system

$△ V$ = Change in volume of the system

Choose the correct answer from the options given below: [2025]

(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
(A)-(II), (B)-(IV), (C)-(III), (D)-(I)

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(3)

(A) Isobaric process, $△ Q = △ U + P △ V$

(B) Isochoric process, $△ Q = △ U$

(C) Adiabatic process, $△ Q = 0$

(D) Isothermal process, $(△ U = 0), △ Q = △ W$

Q 30 :
Match List-I with List-II

List-I List-II

(A) Isothermal (I) $△ W$ (work done) = 0

(B) Adiabatic (II) $△ Q$ (supplied heat) = 0

(C) Isobaric (III) $△ U$ (change in internal energy) $\neq$ 0

(D) Isochoric (IV) $△ U = 0$

Choose the correct answer from the options given below: [2025]

(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
(A)-(IV), (B)-(I), (C)-(III), (D)-(II)
(A)-(IV), (B)-(II), (C)-(III), (D)-(I)
(A)-(II), (B)-(IV), (C)-(I), (D)-(III)

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(3)

For Isothermal,

T = constant, $△ T = 0, △ U = n C_{v} △ T = 0$ ... (iv)

For Adiabatic,

$△ Q = 0$ ... (II)

For Isobaric,

p = constant, $△ U = n C_{v} △ T \neq 0$

$△ Q \neq 0, △ W \neq 0$ ... (III)

For Isochoric,

V = constant, $△ W = 0$ ... (I)

Topic Question Set

Q 21 :
The work done in an adiabatic change in an ideal gas depends upon only: [2025]

Q 25 :
During the melting of a slab of ice at 273 K at atmospheric pressure: [2025]

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	List		List
(A)	Isobaric	(I)	$△ Q = △ W$
(B)	Isochoric	(II)	$△ Q = △ U$
(C)	Adiabatic	(III)	$△ Q = zero$
(D)	Isothermal	(IV)	$△ Q = △ U + P △ V$

	List-I		List-II
(A)	Isothermal	(I)	$△ W$ (work done) = 0
(B)	Adiabatic	(II)	$△ Q$ (supplied heat) = 0
(C)	Isobaric	(III)	$△ U$ (change in internal energy) $\neq$ 0
(D)	Isochoric	(IV)	$△ U = 0$

Topic Question Set

Q 21 : The work done in an adiabatic change in an ideal gas depends upon only: [2025]

Q 25 : During the melting of a slab of ice at 273 K at atmospheric pressure: [2025]

Q 30 : Match List-I with List-II List-I List-II (A) Isothermal (I) △W (work done) = 0 (B) Adiabatic (II) △Q (supplied heat) = 0 (C) Isobaric (III) △U (change in internal energy) ≠ 0 (D) Isochoric (IV) △U=0 Choose the correct answer from the options given below: [2025]

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Q 21 :
The work done in an adiabatic change in an ideal gas depends upon only: [2025]

Q 25 :
During the melting of a slab of ice at 273 K at atmospheric pressure: [2025]