A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between…
A rigid diatomic ideal gas undergoes an adiabatic process at room temperature. The relation between temperature and volume for this process is $\mathrm{TV}^{\mathrm{x}}=$ constant, then $\mathrm{x}$ is:
$\frac{3}{5}$
$\frac{2}{5}$
$\frac{2}{3}$
$\frac{5}{3}$
Solution
Equation of adiabatic change is
$
\mathrm{TV}^{\gamma-1}=\text { constant }
$
Put $\gamma=\frac{7}{5}$, we get: $\gamma-1=\frac{7}{5}-1$
$\therefore \mathrm{x}=\frac{2}{5}$