For an adiabatic process of an ideal gas, $PV^\gamma=\text{constant}$ where $\gamma=\frac{C_P}{C_V}$

We want the relation between $T$ and $V$.

Using the ideal gas law, $PV=nRT \Rightarrow P=\frac{nRT}{V}$

Substitute into the adiabatic equation: $\left(\frac{nRT}{V}\right)V^\gamma=\text{constant}$

$nRT\,V^{\gamma-1}=\text{constant}$

Since $nR$ is constant, $TV^{\gamma-1}=\text{constant}$

Therefore, $x=\gamma-1$

For a rigid diatomic ideal gas at room temperature, vibrational modes are not excited.

So degrees of freedom: $f=5$

Hence, $C_V=\frac{f}{2}R=\frac{5}{2}R$ $C_P=C_V+R=\frac{7}{2}R$

Therefore, $\gamma=\frac{C_P}{C_V} = \frac{\frac{7}{2}R}{\frac{5}{2}R}=\frac{7}{5}$

Now, $x=\gamma-1=\frac{7}{5}-1=\frac{2}{5}$

So the correct option is: $\boxed{\text{B }\left(\frac{2}{5}\right)}$

Comparison with stored answer:

Stored correct answer = B

Our derived answer = B

Hence, they agree.