Thermodynamic cycle on a VT diagram

JEE Advanced 2019 Paper 1, Question 9

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume vs. temperature ( $VT$ ) diagram. The correct statement(s) is/are:

[ $R$ is the gas constant]

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Work done in this thermodynamic cycle ( $1 \to 2 \to 3 \to 4 \to 1$ ) is $|W| = \frac{1}{2} R T_0$ .
The above thermodynamic cycle exhibits only isochoric and adiabatic processes.
The ratio of heat transfer during processes $1 \to 2$ and $2 \to 3$ is $\left| \frac{Q_{1 \to 2}}{Q_{2 \to 3}} \right| = \frac{5}{3}$ .
The ratio of heat transfer during processes $1 \to 2$ and $3 \to 4$ is $\left| \frac{Q_{1 \to 2}}{Q_{3 \to 4}} \right| = \frac{1}{2}$ .

Related article: Thermodynamic processes on an ideal gas

Solution

An ideal gas obeys the relations

(1) $\begin{equation*} \text{The ideal gas law: } P V = n R T \end{equation*}$

(2) $\begin{equation*} \text{Work done {\it by} the gas: } W = \int_a^b P dV \end{equation*}$

(3) $\begin{equation*} \text{Internal energy of a monatomic gas: } U = \frac{3}{2} n R T \end{equation*}$

We are told that there is one mole of gas so $n=1$ . Let us draw a $PV$ diagram for this process.

At point 4 we have $V_4 = V_0$ and, by (1), $P_4 = R T_0/2 V_0$ . From point $4 \to 1$ the volume remains constant, however the temperature increases. That means $P_1 = R T_0/V_0$ .

From point $1 \to 2$ both volume and temperature increase in a linear fashion. That means the pressure must be constant. Therefore $P_2 = P_1$ and $V_2 = 2 V_0$ .

From $2 \to 3$ there is no change in volume but the temperature drops. That means the pressure drops as well. So we have $V_3 = 2V_0$ and by (1), $P_3 = R T_0/2 V_0$ .

Finally, from $3 \to 4$ both volume and temperature decrease in a linear fashion, which means the pressure is a constant. So $P_4 = P_2$ , $V_4 = V_0$ and we have completed the cycle.

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The work done by the gas is the area enclosed by the $PV$ curve,

(4) $\begin{equation*} W_{\rm cycle} = \frac{R T_0}{2 V_0} \times V_0 = \frac{R T_0}{2} \end{equation*}$

So option (A) is correct. It is also clear from the $PV$ diagram that the cycle consists of only isobaric (constant pressure) and isochoric (constant volume) processes. So option (B) is incorrect.

To compute the heat exchanged we recall the first law of thermodynamics,

(5) $\begin{equation*} Q = \Delta U + W . \end{equation*}$

We can compute $\Delta U_{a \to b}$ from the $VT$ diagram using (3), and $W_{a \to b}$ from the $PV$ diagram using (2).

(6) $\begin{gather*} Q_{1 \to 2} = \Delta U_{1 \to 2} + \int_1^2 P dV = \frac{3}{2} R T_0 + R T_0 = \frac{5}{2} R T_0 \\[1em] Q_{2 \to 3} = \Delta U_{2 \to 3} + \int_2^3 P dV = -\frac{3}{2} R T_0 + 0 = -\frac{3}{2} R T_0 \\[1em] Q_{3 \to 4} = \Delta U_{3 \to 4} + \int_3^4 P dV = -\frac{3}{2} R T_0 -\frac{1}{2} R T_0 = -2 R T_0 \end{gather*}$

Therefore,

(7) $\begin{equation*} \left| \frac{Q_{1 \to 2}}{Q_{2 \to 3}} \right| = \frac{5}{3} \end{equation*}$

which means option (C) is correct. However,

(8) $\begin{equation*} \left| \frac{Q_{1 \to 2}}{Q_{3 \to 4}} \right| = \frac{5}{4} \end{equation*}$

and therefore option (D) is incorrect.

The correct answers are (A) and (C).

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