In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by TΔX, where T is temperature of the system and ΔX is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas X = (3/2) R In ((T/TA))+R In ((V/VA)). Here, R is gas constant, V is volume of gas, TA and VA 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→2→3 (p) (1/3)RT0 In 2 (II) Change in internal energy in process 1→2→3 (Q) (1/3)RT0 (III) Heat absorbed by the system in process 1→2→3 (R) RT0 (IV) Heat absorbed by the system in process 1→2 (S) (4/3)RT0 (T) (1/3)RT0 (3+In 2) (U) (5/6)RT0 Question: If the process carried out on one mole of monatomic ideal gas is as shown in figure in the PV-diagram with P0 V0 = (1/3)RT0, the correct match is.

Question

In a thermodynamic process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by TΔX, where T is temperature of the system and ΔX is the infinitesimal change in a thermodynamic quantity X of the system. For a mole of monatomic ideal gas X = (3/2) R In ((T/TA))+R In ((V/VA)). Here, R is gas constant, V is volume of gas, TA and VA 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→2→3 (p) (1/3)RT0 In 2 (II) Change in internal energy in process 1→2→3 (Q) (1/3)RT0 (III) Heat absorbed by the system in process 1→2→3 (R) RT0 (IV) Heat absorbed by the system in process 1→2 (S) (4/3)RT0 (T) (1/3)RT0 (3+In 2) (U) (5/6)RT0 Question: If the process carried out on one mole of monatomic ideal gas is as shown in figure in the PV-diagram with P0 V0 = (1/3)RT0, the correct match is. (A) I → Q, II → R, III → P, IV → U (B) I → Q, II → R, III → S, IV → U (C) I → S, II → R, III → Q, IV → T (D) I → Q, II → S, III → R, IV → U

Tardigrade Admin · Accepted Answer

Degree of freedom f = 3
Work done in any process = Area under P-V graph

\Rightarrow

Work done in

1 \to 2 \to 3 = P_{0} V_{0}

= \frac{R T _{0}}{3} \Rightarrow (Q)

(II) Change in internal energy

1 \to 2 \to 3

Δ U = n C_{v} Δ T

= \frac{f}{2} n R Δ T

= \frac{f}{2} (P_{f} V_{f} - P_{i} V_{i})

= \frac{3}{2} (\frac{3 P _{0}}{2} 2 V_{0} - P_{0} V_{0})

= 3 P_{0} V_{0}

Δ U = R T_{0} \Rightarrow (R)

(III) Heat absorbed in

1 \to 2 \to 3

for any process,

I^{s t}

law of thermodynamics

Δ Q = Δ W + ω

Δ Q = R T_{0} + \frac{R T _{0}}{3}

Δ Q = \frac{4 R T _{0}}{3} \Rightarrow (S)

(IV) Heat absorbed in process

1 \to 2

Δ Q = Δ U + W

= \frac{f}{2} (P_{f} V_{f} - P_{i} V_{i}) + W

= \frac{3}{2} (P_{0} 2 V_{0} - P_{0} V_{0}) + P_{0} V_{0}

= \frac{5}{2} P_{0} V_{0}

= \frac{5}{2} (\frac{R T _{0}}{3})

Δ Q = \frac{5 R T _{0}}{6} \Rightarrow (U)

List-I	List-II
(I) Work done by the system in process 1 $\to$ 2 $\to$ 3	(p) $\frac{1}{3} R T_{0} I n 2$
(II) Change in internal 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 absorbed by the system in process 1 $\to$ 2	(S) $\frac{4}{3} R T_{0}$
	(T) $\frac{1}{3} R T_{0} (3 + I n 2)$
	(U) $\frac{5}{6} R T_{0}$

I $\to$ Q, II $\to$ R, III $\to$ P, IV $\to$ U

I $\to$ Q, II $\to$ R, III $\to$ S, IV $\to$ U

I $\to$ S, II $\to$ R, III $\to$ Q, IV $\to$ T

I $\to$ Q, II $\to$ S, III $\to$ R, IV $\to$ U