Given data

• Number of moles: $n=1$
• Process: isothermal
• Temperature: $27^\circ C = 300\,K$
• Final pressure is double the initial pressure: $P_2 = 2P_1$

Use ideal gas law for isothermal process For an isothermal process of an ideal gas, $P_1V_1 = P_2V_2$ Since $P_2 = 2P_1$, $P_1V_1 = 2P_1V_2$ $V_1 = 2V_2$ $V_2 = \frac{V_1}{2}$

Work done in isothermal compression Work done by the gas in an isothermal process is $W_{\text{by}} = nRT\ln\left(\frac{V_2}{V_1}\right)$ Substituting values: $W_{\text{by}} = (1)R(300)\ln\left(\frac{1}{2}\right)$ $W_{\text{by}} = -300R\ln 2$

Work done on the gas Work done on the gas is the negative of work done by the gas: $W_{\text{on}} = -W_{\text{by}} = 300R\ln 2$

Match with options Therefore, the correct option is: $\boxed{300R\ln 2}$ which is Option C.

Comparison with stored answer Stored correct answer: C

Our derived answer also gives C, so they agree.