Question

For an adiabatic expansion of perfect gas, the value of $\left(\frac{\Delta P}{P}\right)$ is equal to

• A. $-y^{1/2}\left(\frac{\Delta V}{V}\right)$
• B. $-\left(\frac{\Delta V}{V}\right)$
• C. $-\gamma \left(\frac{\Delta V}{V}\right)$
• D. ${\gamma }^2\left(\frac{\Delta V}{V}\right)$

Answer 1

Answer: (C)

$-\gamma \left(\frac{\Delta V}{V}\right)$

In an adiabatic expansion of gas,

$P{V}^{\gamma }=constant$

Taking logarithm on both sides

$ln\left(P\right)+\gamma ln\left(V\right)=ln\left(C\right)$

On differentiating:

$\frac{\Delta P}{P}+\gamma \frac{\Delta V}{V}=0\Rightarrow \frac{\Delta P}{P}=-\gamma \frac{\Delta V}{V}$

Hence, option C is correct.