A work of $166.28 \mathrm{~J}$ is done to adiabatically compress one mole of a gas. If the increase in the…
A work of $166.28 \mathrm{~J}$ is done to adiabatically compress one mole of a gas. If the increase in the temperature of the gas is $8^{\circ} \mathrm{C}$, the gas is
$\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)$
monatomic
diatomic
polyatomic
mixture of diatomic and polyatomic
Solution
Work done on the system, $\mathrm{W}=166.28 \mathrm{~J}$
Increase in temperature, $\Delta \mathrm{T}=\mathrm{T}_2-\mathrm{T}_1=8^{\circ} \mathrm{C}$
Work done in adiabatic process,
$\begin{aligned}
& \mathrm{W}=\frac{\mathrm{nR}\left(\mathrm{T}_2-\mathrm{T}_1\right)}{\gamma-1} \\
& 166.28=\frac{8.314 \times 8}{\gamma-1} \\
& \Rightarrow \gamma-1=\frac{8.314 \times 8}{166.28}=0.4 \\
& \Rightarrow \gamma=1.4
\end{aligned}$
The gas is diatomic in nature.