One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the $P V$ -diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-1 with the corresponding statements in List-II.

List-I List-II

P. In process I 1. Work done by the gas is zero

Q. In process II 2 Temperature of the gas remains unchanged

R. In process III 3 No heat is exchanged between the gas and its surroundings

S. In process IV 4 Work done by the gas is $6 P_{0} V_{0}$

P \to 4; Q \to 3; R \to 1; S \to 2

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P \to 1; Q \to 3; R \to 2; S \to 4

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P \to 3; Q \to 4; R \to 1; S \to 2

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P \to 3; Q \to 4; R \to 2; S \to 1

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Solution

The correct option is C $P \to 3; Q \to 4; R \to 1; S \to 2$
Process 1 is adiabatic process, hence $Q = 0$
Process 2 is isobaric process, hence $W = p . Δ V = 6 P_{0} V_{0}$
Process 3 is isochoric, hence $W = 0$
Process 4 is isothermal

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Q. One mole of a monoatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV- diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List - I with the corresponding statements in List - II.

List - I	List - II
P. In process I	1. Work done by the gas is zero
Q. In process II	2. Temperature of the gas remains unchanged
R. In process III	3. No heat is exchanged between the gas and its surroundings
S. In process IV	4. Work done by the gas is $6 P_{0} V_{0}$

One mole of diatomic gas is taken through a cyclic process. Pressure, volume and temperature of gas at A is $P_{0}$ , $V_{0}$ and $T_{0}$ . List-I gives some physical quantities and List-II gives their possible magnitudes.

List-I

List-II

(1) Work done in process BC	(P) $\frac{3}{2} P_{0} V_{0}$
(2) Work done in process CA	(Q) $2 P_{0} V_{0}$
(3) Change in internal energy during process BC	(R) $3 P_{0} V_{0}$
(4) Change in internal energy during process CA	(S) $5 P_{0} V_{0}$
	(T) $\frac{15}{2} P_{0} V_{0}$
	(U) 0

If cyclic process is given on P-V graph as shown in figure then match List-I with List-II.

In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by $T Δ X$ , where $T$ is temperature of the system and $Δ X$ is the infinitesimal change in a thermodynamic quantity $X$ of the system. For a mole of monatomic ideal gas $X = \frac{3}{2} R ln (\frac{T}{T_{A}}) + Rln (\frac{V}{V_{A}}) .$ Here, $R$ is gas constant, $V$ is volume of gas, $T_{A}$ and $V_{A}$ are constants.

The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

	List-I		List-II
(i)	Work done by the system in process $1 \to 2 \to 3$	(P)	$\frac{1}{3} {R T}_{0} ln 2$
(ii)	Change in intermal energy in process $1 \to 2 \to 3$	(Q)	$\frac{1}{3} {R T}_{0}$
(iii)	Heat absorbed by the system in process $1 \to 2 \to 3$	(R)	${R T}_{0}$
(iv)	Heat absotbed by the system in process $1 \to 2$	(S)	$\frac{4}{3} {R T}_{0}$
		(T)	$\frac{1}{3} {R T}_{0} (3 + ln 2)$
		(U)	$\frac{5}{6} {R T}_{0}$

If the process carricd out on one mole of monoatomic ideal gas is as shown in figure in the PV-diagram with $P_{0} V_{0} = \frac{1}{3} {R T}_{0}$ , the correct match is,

One mole of diatomic gas is taken through a cyclic process. Pressure, volume and temperature of gas at A is $P_{0}, V_{0}$ and $T_{0}$ . List-I gives some physical quantities and List-II gives their possible magnitudes.

List-I

List-II

(1) Work done in process BC	(P) $\frac{3}{2} P_{0} V_{0}$
(2) Work done in process CA	(Q) $2 P_{0} V_{0}$
(3) Change in internal energy during process BC	(R) $3 P_{0} V_{0}$
(4) Change in internal energy during process CA	(S) $5 P_{0} V_{0}$
	(T) $\frac{15}{2} P_{0} V_{0}$
	(U) 0

If cyclic process is given on P-T graph as shown in figure. Then match List-I with List-II.

Q. The figures given below show different processes (relating pressure

P

and volume

V

) for a given amount for an ideal gas.

Δ W

is work done by the gas and

Δ Q

is heat absorbed by the gas.

\begin{matrix} Column I & Column II 1. In Fig (i) & (p) Δ Q > 0 2. In Fig (i i) & (q) Δ W < 0 3. In Fig (i i i) & (r) Δ Q < 0 4. In Fig (i v) & (s) Δ W > 0 \end{matrix}

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The correct option is C P→3;Q→4;R→1;S→2Process 1 is adiabatic process, hence Q=0 Process 2 is isobaric process, hence W=p.ΔV=6P0V0 Process 3 is isochoric, hence W=0 Process 4 is isothermal

The correct option is C $P \to 3; Q \to 4; R \to 1; S \to 2$
Process 1 is adiabatic process, hence $Q = 0$
Process 2 is isobaric process, hence $W = p . Δ V = 6 P_{0} V_{0}$
Process 3 is isochoric, hence $W = 0$
Process 4 is isothermal