Air compression

Isothermal Compression: Since the temperature $T = T_0$ is constant, the ideal gas follows Boyle's law $\begin{aligned} & & p V &= n R T_0 = \text{const}\\ \Rightarrow & & p &= \frac{n R T_0}{V} \propto V^{-1} \\ \Rightarrow & & \Delta W &= - \int_{V_0}^{V_0/2} p dV = - n R T_0 \int_{V_0}^{V_0/2} \frac{dV }{V} \\ & & &= - n R T_0 (\log(V_0/2) - \log(V_0)) = \log(2) \cdot n R T_0 \\ & & &\approx 0.69 \cdot n R T_0 \end{aligned}$ with the molar gas quantity, $n$ , and the gas constant, $R$ .

Adiabatic-Isobaric Compression: In adiabatic process, no heat is exchanged with the environment. The 1st law of thermodynamics yields $\begin{aligned} & & \delta U &= \underbrace{\delta Q}_{=0} + \delta W = \delta W, \quad U = \frac{f}{2} n R T = \frac{f}{2} p V \\ \Rightarrow & & \delta U &= \frac{f}{2} \left( V dp + p dV \right) = - p dV = \delta W \\ \Rightarrow & & \frac{dp}{p} &= - \frac{f + 2}{f} \frac{dV}{V} \\ \Rightarrow & & \int_{p_0}^{p} \frac{dp}{p} &= - \frac{f + 2}{f} \int_{V_0}^{V} \frac{dV}{V} \\ \Rightarrow & & \log \frac{p}{p_0} &= - \frac{f + 2}{f} \log \frac{V}{V_0} \\ \Rightarrow & & p &= p_0 \left( \frac{V_0}{V} \right)^\gamma \propto V^{-\gamma}, \quad \gamma = \frac{f+2}{f} \end{aligned}$ with the degrees of freedom, $f$ , and the adiabatic exponent, $\gamma$ . For air at room temperature these constants yield $f = 5$ and $\gamma = 1.4$ , since air consists of diatomic molecules, which have three degrees of freedom of translation and two degrees of freedom of rotation. When the pressure is doubled, we obtain a final volume $V_1$ for the adiabatic compression with: $\begin{aligned} & & p &= 2 p_0 = p_0 \left( \frac{V_0}{V_1} \right)^\gamma \\ \Rightarrow & & V_1 &= 2^{-1/\gamma} V_0 \approx 0.61 V_0 \end{aligned}$ The final state with $V = 0.5 V_0$ is achieved by isobaric compression with $p = 2 p_0 = \text{const}$ . The total work yields $\begin{aligned} \Delta W &= - \int_{V_0}^{V_1} p dV - \int_{V_1}^{V_0/2} p dV \\ &= - p_0 V_0^\gamma \int_{V_0}^{V_1} \frac{dV}{V^\gamma} - 2 p_0 \int_{V_1}^{V_0/2} dV \\ &= \frac{1}{ \gamma - 1} p_0 V_0^\gamma \left( \frac{1}{V_1^{\gamma-1}} - \frac{1}{V_0^{\gamma- 1}} \right) - 2 p_0 \left( \frac{V_0}{2} - V_1 \right)\\ &= \frac{2 \gamma - 1}{\gamma -1} \left( 2^{(\gamma - 1)/\gamma} - 1 \right) p_0 V_0 = \frac{2 \gamma - 1}{\gamma -1} \left( 2^{(\gamma - 1)/\gamma} - 1 \right) n R T_0 \\ &\approx 0.99 \cdot n R T_0 \end{aligned}$ Since this work is greater, it is better to compress the gas isothermally.