Question

The variation of pressure $P$ with volume $V$ for an ideal monoatomic gas during an adiabatic process is shown in figure. At point $A$, find the magnitude of the rate of change of pressure with volume.

• A. $\frac{3P_0}{5V_0}$
• B. $\frac{5P_0}{3V_0}$
• C. $\frac{3P_0}{2V_0}$
• D. $\frac{5P_0}{2V_0}$

Answer 1

The correct option is D $\frac{5P_0}{2V_0}$

We can find the adiabatic coefficient $\gamma$ by using,
$\gamma =1+\frac{2}{f}$
Since the gas is monoatomic in nature,
Number of degrees of freedom $\left(f\right)=3$
Thus we get, $\gamma =\frac{5}{3}$

Equation of state of an adiabatic process in terms of $P$ and $V$ is given by
$P{V}^{\gamma }=$ constant
Differentiating both sides,
$\frac{dP}{P}=-\gamma \frac{dV}{V}$
We can write this as,
$\frac{dP}{dV}=-\gamma \times \frac{P}{V}$
At point $A$, from the data given in the figure, we can say that
$\left|\frac{dP}{dV}\right|=\frac{5}{3}\times \frac{3P_0}{2V_0}=\frac{5P_0}{2V_0}$
Thus, option (d) is the correct answer.