Question

A diatomic gas is compressed adiabatically to $\frac{1}{32}$ of its initial volume. If the initial temperature of gas is $T_{1}K$ and final is $T_2=a{T}_{1}K$, then value of $a$ is

• A. 4
• B. 6
• C. 5
• D. 9

Answer 1

The correct option is A 4

$\gamma =\frac{7}{5}$ for diatomic gas.
For adiabatic process,
$\frac{T_2}{T_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma -1}$
$\frac{T_2}{T_1}=(32{)}^{\frac{7}{5}-1}=2^{5\times \frac{2}{5}}=4$
$T_2=4T_1$