Question

A gas consisting of rigid diatomic molecules was initially under standard condition, then gas was compressed adiabatically to one-fifth of its initial volume. What will be the mean kinetic energy of a rotating molecule in the final state?
$\left[\right.$ Take, $\gamma$ for diatomic gas $\left.=\frac{7}{5}\right]$

• A. $1.44\, J$
• B. $4.55\, J$
• C. $787.98 \times 10^{-23} J$
• D. $757.3 \times 10^{-23} J$

Answer 1

Answer: (C)

Given, for diatomic gas, $\gamma=\frac{7}{5}$
For adiabatic process, $T V^{\gamma-1}=$ constant
$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma-1}$
Given $V_{2}=\frac{V_{1}}{5} \text { and } T_{1}=27^{\circ} C =300\, K$
(at standard conditions)
$(300) V_{1}^{\frac{7}{5}-1}=T_{2}\left(\frac{V_{1}}{5}\right)^{\frac{7}{5}-1}$
$\Rightarrow T_{2}=\frac{300 \times V_{1}^{2 / 5}}{V_{1}^{\frac{2}{5}} \times\left(\frac{1}{5}\right)^{2 / 5}}=\frac{300}{5^{\frac{-2}{5}}}=300 \times 5^{2 / 5}$
$=300 \times 1.903=571 \,K$
Mean kinetic energy of rotating molecules
$=k T=1.38 \times 10^{-23} \times 571=787.98 \times 10^{-23} \,J$