Question:
A quasi-static cycle of a monoatomic ideal gas contains an isothermal process \((ab)\), followed by an isochoric process \((bc)\) and an adiabatic process \((ca)\) as shown in the figure. The volumes of the gas are \(V_1\) and \(V_2\) at \(a\) and \(b\), respectively. If the cycle has heat input \(Q_{\mathrm{in}}\) and output \(Q_{\mathrm{out}}\), then the efficiency of the cycle is defined as
\[
\eta=\frac{Q_{\mathrm{in}}-Q_{\mathrm{out}}}{Q_{\mathrm{in}}}
\]
The correct statement(s) is/are:
\[
[\text{Given: }\ln2\approx0.7]
\]
A quasi-static cycle of a monoatomic ideal gas contains an isothermal process \((ab)\), followed by an isochoric process \((bc)\) and an adiabatic process \((ca)\) as shown in the figure. The volumes of the gas are \(V_1\) and \(V_2\) at \(a\) and \(b\), respectively. If the cycle has heat input \(Q_{\mathrm{in}}\) and output \(Q_{\mathrm{out}}\), then the efficiency of the cycle is defined as
\[
\eta=\frac{Q_{\mathrm{in}}-Q_{\mathrm{out}}}{Q_{\mathrm{in}}}
\]
The correct statement(s) is/are:
\[
[\text{Given: }\ln2\approx0.7]
\]
Show Hint
For adiabatic process:
\[
TV^{\gamma-1}=\text{constant}
\]
For isothermal process:
\[
PV=\text{constant}
\]
Updated On: May 20, 2026
- If \(\dfrac{V_2}{V_1}=8\), the heat released in process \(bc\) is smaller than the heat absorbed in process \(ab\)
- For a given value of \(\dfrac{V_2}{V_1}\), \(\eta\) does not depend on the temperature of the isothermal process
- If \(\dfrac{V_2}{V_1}=8\), then temperature at \(a\) is \(4\) times temperature at \(c\)
- If \(\dfrac{V_2}{V_1}=8\), then pressure at \(a\) is \(4\) times pressure at \(b\)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Analyze the processes.
Process: \[ ab \] is isothermal. Hence: \[ T_a=T_b \] Process: \[ bc \] is isochoric. Process: \[ ca \] is adiabatic. For monoatomic gas: \[ \gamma=\frac53 \]
Step 2: Use adiabatic relation between \(c\) and \(a\).
For adiabatic process: \[ TV^{\gamma-1}=\text{constant} \] Thus: \[ T_cV_2^{2/3}=T_aV_1^{2/3} \] \[ \frac{T_a}{T_c} = \left(\frac{V_2}{V_1}\right)^{2/3} \] If: \[ \frac{V_2}{V_1}=8 \] \[ \frac{T_a}{T_c}=8^{2/3}=4 \] Hence: \[ \boxed{\mathrm{(C)\ is\ correct}} \]
Step 3: Find pressure ratio for isothermal process.
For isothermal process: \[ PV=\text{constant} \] Thus: \[ P_aV_1=P_bV_2 \] \[ \frac{P_a}{P_b} = \frac{V_2}{V_1} \] If: \[ \frac{V_2}{V_1}=8 \] \[ \frac{P_a}{P_b}=8 \] Hence statement: \[ P_a=4P_b \] is false. Therefore: \[ \boxed{\mathrm{(D)\ is\ incorrect}} \]
Step 4: Compare heats in processes \(ab\) and \(bc\).
Heat absorbed in isothermal expansion: \[ Q_{ab}=nRT_a\ln\left(\frac{V_2}{V_1}\right) \] For: \[ \frac{V_2}{V_1}=8 \] \[ Q_{ab}=nRT_a\ln8 \] \[ =3nRT_a\ln2 \] Using: \[ \ln2\approx0.7 \] \[ Q_{ab}\approx2.1nRT_a \] Now for isochoric process: \[ Q_{bc}=nC_V(T_c-T_b) \] Since: \[ T_b=T_a,\qquad T_c=\frac{T_a}{4} \] \[ Q_{bc} = n\left(\frac32R\right)\left(\frac{T_a}{4}-T_a\right) \] \[ = -\frac98nRT_a \] Magnitude: \[ |Q_{bc}|=1.125\,nRT_a \] Thus: \[ |Q_{bc}|<Q_{ab} \] Hence: \[ \boxed{\mathrm{(A)\ is\ correct}} \]
Step 5: Check efficiency dependence.
Efficiency: \[ \eta = 1-\frac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}} \] Both heats are proportional to: \[ T_a \] Hence temperature cancels out. Thus efficiency depends only on: \[ \frac{V_2}{V_1} \] Therefore: \[ \boxed{\mathrm{(B)\ is\ correct}} \]
Step 6: Identify correct statements.
Therefore: \[ \boxed{\mathrm{(A),\ (B)\ and\ (C)}} \]
Process: \[ ab \] is isothermal. Hence: \[ T_a=T_b \] Process: \[ bc \] is isochoric. Process: \[ ca \] is adiabatic. For monoatomic gas: \[ \gamma=\frac53 \]
Step 2: Use adiabatic relation between \(c\) and \(a\).
For adiabatic process: \[ TV^{\gamma-1}=\text{constant} \] Thus: \[ T_cV_2^{2/3}=T_aV_1^{2/3} \] \[ \frac{T_a}{T_c} = \left(\frac{V_2}{V_1}\right)^{2/3} \] If: \[ \frac{V_2}{V_1}=8 \] \[ \frac{T_a}{T_c}=8^{2/3}=4 \] Hence: \[ \boxed{\mathrm{(C)\ is\ correct}} \]
Step 3: Find pressure ratio for isothermal process.
For isothermal process: \[ PV=\text{constant} \] Thus: \[ P_aV_1=P_bV_2 \] \[ \frac{P_a}{P_b} = \frac{V_2}{V_1} \] If: \[ \frac{V_2}{V_1}=8 \] \[ \frac{P_a}{P_b}=8 \] Hence statement: \[ P_a=4P_b \] is false. Therefore: \[ \boxed{\mathrm{(D)\ is\ incorrect}} \]
Step 4: Compare heats in processes \(ab\) and \(bc\).
Heat absorbed in isothermal expansion: \[ Q_{ab}=nRT_a\ln\left(\frac{V_2}{V_1}\right) \] For: \[ \frac{V_2}{V_1}=8 \] \[ Q_{ab}=nRT_a\ln8 \] \[ =3nRT_a\ln2 \] Using: \[ \ln2\approx0.7 \] \[ Q_{ab}\approx2.1nRT_a \] Now for isochoric process: \[ Q_{bc}=nC_V(T_c-T_b) \] Since: \[ T_b=T_a,\qquad T_c=\frac{T_a}{4} \] \[ Q_{bc} = n\left(\frac32R\right)\left(\frac{T_a}{4}-T_a\right) \] \[ = -\frac98nRT_a \] Magnitude: \[ |Q_{bc}|=1.125\,nRT_a \] Thus: \[ |Q_{bc}|<Q_{ab} \] Hence: \[ \boxed{\mathrm{(A)\ is\ correct}} \]
Step 5: Check efficiency dependence.
Efficiency: \[ \eta = 1-\frac{Q_{\mathrm{out}}}{Q_{\mathrm{in}}} \] Both heats are proportional to: \[ T_a \] Hence temperature cancels out. Thus efficiency depends only on: \[ \frac{V_2}{V_1} \] Therefore: \[ \boxed{\mathrm{(B)\ is\ correct}} \]
Step 6: Identify correct statements.
Therefore: \[ \boxed{\mathrm{(A),\ (B)\ and\ (C)}} \]
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