Question

For a reversible adiabatic expansion of an ideal gas $dp/p$ equal to:

• A. $\gamma \frac{dv}{v}$
• B. $\frac{dv}{v}$
• C. $\left(\frac{\gamma }{\gamma -1}\right)\frac{dv}{v}$
• D. $-\gamma \frac{dv}{v}$

Answer 1

The correct option is D $-\gamma \frac{dv}{v}$

Solution:- (D) $-\gamma \frac{dv}{v}$

In an adiabatic process,

$p{v}^{\gamma }=K\left(\text{constant}\right)$

Differentiating the above equation, we have

$\left(1.dp\right)v^{\gamma }+p.\left(\gamma v^{\gamma -1}dv\right)=0$

$dp{v}^{\gamma }=-\gamma p{v}^{\gamma -1}dv$

$\frac{dp}{p}=\frac{-\gamma v^{\gamma -1}}{v^{\gamma }}dv$

$\frac{dp}{p}=-\gamma \frac{dv}{v}$

Here $\frac{dp}{p}$ and $\frac{dv}{v}$ are the fractional change in pressure and volume respectively.