The ratio of the slopes of isothermal and adiabatic curves is
The ratio of the slopes of isothermal and adiabatic curves is
1
\(\gamma\)
\(\frac{1}{\gamma}\)
\(\frac{3}{2}\)
Solution
For an isothermal change,
\(p V=K\)
Differentiating both side, we get
\(p \cdot d V+V d p=0 \Rightarrow V d p=-p d V\)
\(\therefore\) Slope of an isothermal curve
\(\left(\frac{d p}{d V}\right)_{\text {iso }}=-\frac{p}{V}\)
For an adiabatic change,
\(p V^\gamma=K^{\prime}\)
Differentiating both side, we get
\(\begin{aligned}
& p \cdot \gamma V^{\gamma-1} \cdot d V+V^\gamma \cdot d p=0 \\
& \Rightarrow \quad \gamma p d V+V d p=0 \text { or } \quad V d p=-\gamma \cdot p d V \\
\end{aligned}\)
\(\therefore\) Slope of an adiabatic curve,
\(\begin{aligned}
& \left(\frac{d p}{d V}\right)_{\text {adia }}=-\frac{\gamma p}{V} \\
\therefore \quad & \frac{\left(\frac{d p}{d V}\right)_{\text {iso }}}{\left(\frac{d p}{d V}\right)_{\text {adia }}}=\frac{-p / V}{-\gamma p / V}=\frac{1}{\gamma}
\end{aligned}\)