Question:
The slopes of the isothermal and adiabatic \(p-v\) graphs of a gas are by \(S_I\) and \(S_A\) respectively. If the heat capacity ratio of the gas is \(\dfrac{3}{2}\), then
\[
\frac{S_I}{S_A}=
\]
The slopes of the isothermal and adiabatic \(p-v\) graphs of a gas are by \(S_I\) and \(S_A\) respectively. If the heat capacity ratio of the gas is \(\dfrac{3}{2}\), then
\[
\frac{S_I}{S_A}=
\]
Show Hint
The adiabatic curve is steeper than the isothermal curve because
\[
S_A=\gamma S_I
\]
where \(\gamma\gt 1\).
Updated On: Jun 22, 2026
- \(\dfrac{3}{2}\)
- \(\dfrac{2}{3}\)
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{3}\)
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Slope of isothermal curve.
For an isothermal process, \[ PV=\text{constant} \] Differentiating, \[ P\,dV+V\,dP=0 \] Thus, \[ \left(\frac{dP}{dV}\right)_I = -\frac{P}{V} \] Hence, \[ S_I=-\frac{P}{V} \]
Step 2: Slope of adiabatic curve.
For an adiabatic process, \[ PV^\gamma=\text{constant} \] Differentiating, \[ V^\gamma dP+\gamma PV^{\gamma-1}dV=0 \] \[ \frac{dP}{dV} = -\gamma\frac{P}{V} \] Hence, \[ S_A=-\gamma\frac{P}{V} \]
Step 3: Find the ratio of slopes.
\[ \frac{S_I}{S_A} = \frac{-\frac{P}{V}} {-\gamma\frac{P}{V}} \] \[ \frac{S_I}{S_A} = \frac{1}{\gamma} \] Given, \[ \gamma=\frac{3}{2} \] Therefore, \[ \frac{S_I}{S_A} = \frac{1}{3/2} \] \[ \frac{S_I}{S_A} = \frac{2}{3} \]
Step 4: Final conclusion.
Hence, \[ \boxed{\frac{2}{3}} \]
For an isothermal process, \[ PV=\text{constant} \] Differentiating, \[ P\,dV+V\,dP=0 \] Thus, \[ \left(\frac{dP}{dV}\right)_I = -\frac{P}{V} \] Hence, \[ S_I=-\frac{P}{V} \]
Step 2: Slope of adiabatic curve.
For an adiabatic process, \[ PV^\gamma=\text{constant} \] Differentiating, \[ V^\gamma dP+\gamma PV^{\gamma-1}dV=0 \] \[ \frac{dP}{dV} = -\gamma\frac{P}{V} \] Hence, \[ S_A=-\gamma\frac{P}{V} \]
Step 3: Find the ratio of slopes.
\[ \frac{S_I}{S_A} = \frac{-\frac{P}{V}} {-\gamma\frac{P}{V}} \] \[ \frac{S_I}{S_A} = \frac{1}{\gamma} \] Given, \[ \gamma=\frac{3}{2} \] Therefore, \[ \frac{S_I}{S_A} = \frac{1}{3/2} \] \[ \frac{S_I}{S_A} = \frac{2}{3} \]
Step 4: Final conclusion.
Hence, \[ \boxed{\frac{2}{3}} \]
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