Question

If $\gamma$ denotes the ratio of two specific heats of a gas, the ratio of slopes of adiabatic and isothermal PV curves at their point of intersection is

• A. $\frac{1}{\gamma }$
• B. $\gamma$
• C. $\gamma -1$
• D. $\gamma +1$

Answer 1

The correct option is B $\gamma$

For isothermal process $T$ is constant. For $n$ mol of gas we can say PV = constant.
Differentiating both side we get $PdV+VdP=0$
Slope $\left(m_1\right)$ = $\left(\frac{dP}{dV}\right)=\frac{-P}{V}$ ....(i)
For Adiabtic process , according to Poisson equation
$P{V}^{\gamma }$ = constant
Differentiating both side we get $\gamma P{V}^{\gamma -1}dV+V^{\gamma }dP=0$
Slope $\left(m_2\right)$ = $\left(\frac{dP}{dV}\right)=\frac{-\gamma P}{V}$
Ratio of slopes = $\frac{m_2}{m_1}=\frac{\left(\frac{-\gamma P}{V}\right)}{\left(\frac{-P}{V}\right)}$
Ratio of slopes = $\gamma$
Slope of adiabatic curve = $\gamma$ (Slope of isothermal curve)