Thermodynamics

Assuming that the fluorocarbon gas behaves as an ideal gas, then it follows:

$\begin{cases} VT^{\frac{C_V}{R}} = \text{constant} & ...(1)^\circ_{rev} \\ pV^\gamma = \text{constant} & ...(2)^\circ_{rev} \end{cases}$

$\begin{array} {ll} \quad \text{where} & R = 8.3145 \text{ J mol}^{-1} \text{ K}^{-1} \text{, the universal gas constant} \\ & \gamma = \dfrac{C_P}{C_V} \\ & C_P = \text{heat capacity at constant pressure} \\ & C_V = \text{heat capacity at constant volume} \end{array}$

$\begin{aligned} (1): \quad V_iT_i^{\frac{C_V}{R}} & = V_fT_f^{\frac{C_V}{R}} \\ T_i^{\frac{C_V}{R}} & = 2T_f^{\frac{C_V}{R}} \\ \Rightarrow C_V & = \frac{R\log 2}{\log \left(\frac{T_i}{T_f}\right)} \\ (2): \quad p_iV_i^\gamma & = p_fV_f^\gamma \\ p_i & = 2^\gamma p_f \\ \Rightarrow \gamma & = \frac{\log \left(\frac{p_i}{p_f}\right)}{\log 2} \\ \Rightarrow C_P & = \frac{\log \left(\frac{p_i}{p_f}\right)C_V }{\log 2} = \frac{R \log \left(\frac{p_i}{p_f}\right)}{\log \left(\frac{T_i}{T_f}\right)} \\ & = \frac{8.3145 \log \left(\frac{202.94}{81.840}\right)}{\log \left(\frac{298.15}{248.44}\right)} = \boxed{41.40} \text{ J mol}^{-1} \text{ K}^{-1} \end{aligned}$