One mole of an ideal diatomic gas expands from volume $V$ to 2 V isothermally at a temperature $27^{

Question

One mole of an ideal diatomic gas expands from volume $V$ to 2 V isothermally at a temperature $27^{\circ} \mathrm{C}$ and does $W$ joule of work. If the gas undergoes same magnitude of expansion adiabatically from $27^{\circ} \mathrm{C}$ doing the same amount of work $W$, then its final temperature will be (close to) $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.
( $\log _{\mathrm{e}} 2=0.693$ )

Professor PC Thomas Classes · Answer

For isothermal expansion, work done is $W = nRT \ln(\frac{V_2}{V_1}) = 1 	imes R 	imes 300 	imes \ln(2) = 300 R (0.693) = 207.9 R$. For an adiabatic process, work done is $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$. For a diatomic gas, $\gamma = \frac{7}{5} = 1.4$. Equating the work done: $207.9 R = \frac{1 	imes R 	imes (300 - T_2)}{1.4 - 1} = \frac{R(300 - T_2)}{0.4}$. Solving for $T_2$: $300 - T_2 = 207.9 	imes 0.4 = 83.16 \implies T_2 = 300 - 83.16 = 216.84 	ext{ K}$. Converting to Celsius: $216.84 - 273.15 = -56.31^\circ	ext{C}$. Thus, it is close to $-56^\circ	ext{C}$.