Question

The coefficient of volume expansion of an ideal diatomic gas, for a reversible adiabatic process is

• A. $-\frac{5}{7T}$
• B. $-\frac{5}{2T}$
• C. $\frac{7}{2T}$
• D. $\frac{2}{5T}$

Answer 1

The correct option is B $-\frac{5}{2T}$

In any process the coefficient of volume expansion, $r=\frac{1}{V}\frac{dV}{dT}$.

For reversible adiabatic process, we know
$\Rightarrow T{V}^{\gamma -1}=$ constant

Taking $log$ on both sides
$\Rightarrow log\left(T\right)+(\gamma -1)log\left(V\right)=log\left(constant\right)$

Now differentiate both sides, we get
$\Rightarrow \frac{dT}{T}+(\gamma -1)\frac{dV}{V}=0$
$\Rightarrow \frac{dV}{V}=-\frac{1}{(\gamma -1)}\frac{dT}{T}$
$\Rightarrow \frac{1}{V}\frac{dV}{dT}=-\frac{1}{(\gamma -1)T}$
$\Rightarrow r=-\frac{1}{(\gamma -1)T}(\because \gamma =\frac{7}{5})\Rightarrow r=-\frac{5}{2T}.$