Thermodynamics Part 3

Given, Change in pressure with volume $\frac{dp}{dV} = -a p$ At $V=0, p=p_{0}$ Number of mole n = 1 Gas constant, $R=8.314 \mathrm{J} \mathrm{K}^{-1}$ From given equation, $\frac{dp}{p} = -a dV$ $\int\limits_{p_0}^{p} \frac{dp}{p} = -a \int\limits_{0}^{v} dV$ $\Rightarrow [\ln p]_{p_0}^{p} = -a [V]_{0}^{V}$ $\Rightarrow \ln\left( \frac{p}{p_0} \right) = -a (V-0)$ Taking anti log on both side, $\Rightarrow \frac{p}{p_0} = e^{-aV}$ $\Rightarrow p = p_0 e^{-aV}$ By using ideal gas law, pV = nRT $\Rightarrow (p_0 e^{-aV}) V = nRT$ $\Rightarrow \frac{1}{nR} (p_0 e^{-aV}) V = T$ $\Rightarrow \frac{1}{R} (p_0 e^{-aV}) V = T$ [∵ n = 1]…(i) For maximum temperature $(T_{\max})$, On differentiating both side with respect to V, we get $\Rightarrow \frac{1}{nR} \left[ p_0 \cdot e^{-aV} \cdot (-aV) V + p_0 e^{-aV} \right] = \frac{dT}{dV} = 0$ $\Rightarrow p_0 e^{-aV}(1-a V^2) = 0$ $\Rightarrow 1 = a V^2$ $\Rightarrow V^2 = \frac{1}{a}$ $\Rightarrow V = \frac{1}{a}$ Substituting the value in Eq. (i), we get $T = \frac{1}{R} \left(p_0 e^{-a \cdot \frac{1}{a}}\right) \frac{1}{a} = \frac{p_0 e^{-1}}{R} \left( \frac{1}{a} \right) = \frac{p_0}{e R a}$