Question

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume vs. temperature (V-T) diagram. The correct statement(s) is/are [$R$ is the gas constant]

• A. The above thermodynamic cycle exhibits only isochoric and adiabatic processes
• B. The ratio of heat transfer during processes $1\to 2$ and $3\to 4$ is $\left|\frac{Q_{1-2}}{Q_3-4}\right|=\frac{1}{2}$
• C. The ratio of heat transfer during processes $1\to 2$ and $2\to 3$ is $\left|\frac{Q_{1-2}}{Q_{2-3}}\right|=\frac{5}{3}$
• D. Work done in this themodynamic cycle $(1\to 2\to 3\to 4\to 1)$ is $|W|=\frac{1}{2}R{T}_0$

Answer 1

Answer: (C), (D)

Work done in this themodynamic cycle $(1\to 2\to 3\to 4\to 1)$ is $|W|=\frac{1}{2}R{T}_0$

Let the variable at point $1$ is $(P_0,V_0,T_0)$.

Because process $1-2$ is isobaric process so at point 2, $(P_0,2V_0,2T_0)$.

From process 2-3 volume is constant so, it calls an isochoric process. $(\frac{P_0}{2},2V_0,T_0)$.

From process 4-3 is a isobaric proces. So $(\frac{P_0}{2},V_0,\frac{T_0}{2})$

P-V digram,

For process $1-2$ heat transfer,
$Q_{1-2}=n{C}_p{T}_0$
For process $2-3$ heat transfer,
$Q_{2-3}=n{C}_v.(-T_0)$
From $C_p-C_v=R$
and for monoatomic gass $C_v=\frac{3R}{2}{C}_p=\frac{5R}{2}$
so the ratio of $\frac{Q_{1-2}}{Q_{2-3}}=\frac{C_p}{C_v}=\frac{5}{3}$
Hence option C is correct.

Here, option A is wrong because no adiabatic process involved in this process.

Calculation for total work done,
$W_{total}=\frac{P_0}{2}{V}_o=\frac{1}{2}R{T}_0$
Hence, option D is correct.

For option B,
process (3-4) isobaric process,
where, $Q_{1-2}=n{C}_p{T}_0{Q}_{3-4}=n{C}_p\frac{T_0}{2}$
hence, $\frac{Q_{1-2}}{Q_{3-4}}=\frac{2}{1}$
Hence, option B is also wrong.