Two moles of an ideal diatomic gas ($\gamma = 7/5$) at an initial temperature of $T \mathrm{~K}$ do

Question

Two moles of an ideal diatomic gas ($\gamma = 7/5$) at an initial temperature of $T \mathrm{~K}$ do $15R$ joules of work adiabatically. The final temperature of the gas will be

Professor PC Thomas Classes · Answer

(d) : The work done by an ideal gas in an adiabatic process is given by $W = \frac{nR(T_i - T_f)}{\gamma - 1}$. Substituting the given values: $15R = \frac{2R(T - T_f)}{(7/5) - 1}$. This simplifies to $15 = \frac{2(T - T_f)}{2/5} = 5(T - T_f)$. Dividing both sides by 5 gives $3 = T - T_f$. Therefore, the final temperature is $T_f = (T - 3) \mathrm{~K}$.