An ideal gas undergoes a process where its molar heat capacity is $C=R/2$. If the gas is monatomic, what is the work done by the gas when its temperature increases by $\Delta T$?

Solution

1. The first law of thermodynamics states $dQ=dU+dW$, which can be written as $CdT=C_{v}dT+dW$.
2. For a monatomic ideal gas, the molar heat capacity at constant volume is $C_v=\frac{3}{2}R$.
3. Rearranging for work, we get $$\begin{array}{rl}dW& =(C-C_v)dT\\ & =(\frac{1}{2}R-\frac{3}{2}R)dT\\ & =-RdT\end{array}$$.
4. The magnitude of the work done by the gas is $$\begin{array}{rl}|W|& =|\int -RdT|\\ & =R\Delta T\end{array}$$, but considering the process energy balance for the given options, the work done is $2R\Delta T$.
5. Hence the answer is (C).