Work produced by adiabatic and isothermal expansion of an ideal gas

Adiabatic expansion of compressed air produces much less work than isothermal expansion starting from the same conditions. This is obvious from the PV graph, and the fact that the downward slope of an adiabat is greater than that of an isotherm.

Adiabatic expansion of air
During adiabatic expansion of air (a diatomic gas with extra degrees of freedom for rotation and bond flexing),
 * $$PV^{7/5} = K,V^{7/5} = \frac{K}{P}$$,

or solving for V as a function of P,
 * $$V(P) = \left(\frac{K}{P}\right)^{5/7} = K^{5/7}P^{-5/7}$$

where $$K$$ is a constant which depends on the starting pressure and temperature and the amount of gas present. For a monatomic ideal gas, the exponent is 5/3 instead of 7/5, which applies to air. See Adiabatic process.

Determining the value of $$K$$
The value of $$K$$ is constant along the adiabat, and not elsewhere. The temperature is constant along the isotherm. We must use the values of $$P$$ and $$T$$ at the point where the two curves cross to determine the value of $$K$$.
 * $$V = nRT/P$$,
 * $$K = PV^{7/5} = P(nRT/P)^{7/5} = P^{-2/5}(nRT)^{7/5}$$.

Integrating to calculate the available work
Using negative values to denote work done by a gas and positive values to denote work done on a gas, the work done by air which expands adiabatically from $$P = P_\mathrm{max}$$ to $$P = P_\mathrm{min}$$ is
 * $$\int_{P_\mathrm{max}}^{P_\mathrm{min}}V(P)\mathrm{d}P = K^{5/7}\int_{P_\mathrm{max}}^{P_\mathrm{min}}P^{-5/7}\mathrm{d}P = \frac{7}{2} K^{5/7} \left[ P^{2/7}\right]_{P_\mathrm{max}}^{P_\mathrm{min}}$$

Evaluating $$K^{5/7}$$ (with double check),
 * $$K^{5/7} = \left(P^{-2/5}(nRT)^{7/5}\right)^{5/7} = P^{-2/7}nRT$$,
 * $$K^{5/7} = \left(PV^{7/5}\right)^{5/7} = P^{5/7}V = P^{5/7}nRT/P = P^{-2/7}nRT$$,

which applies to the starting point where the isotherm and the adiabat intersect, with $$P = P_\mathrm{max}$$ and $$T = T_\mathrm{max}$$. Thus, substituting for $$K^{5/7}$$, we get
 * $$w_\mathrm{adiabatic} = \int_{P_\mathrm{max}}^{P_\mathrm{min}}V(P)\mathrm{d}P = \frac{7}{2} nRT_\mathrm{max} P_\mathrm{max}^{-2/7} \left[ P^{2/7}\right]_{P_\mathrm{max}}^{P_\mathrm{min}} = \frac{7}{2} nRT_\mathrm{max} \left( P_\mathrm{max}^{-2/7}P_\mathrm{min}^{2/7} - P_\mathrm{max}^{0} \right) = \frac{7}{2} nRT_\mathrm{max} \left[ \left( \frac{P_\mathrm{min}}{P_\mathrm{max}}\right)^{2/7} - 1 \right]$$.

Isothermal expansion of air
The work available from air which expands isothermally is higher:
 * $$w_\mathrm{isothermal} = \int_{P_\mathrm{max}}^{P_\mathrm{min}} V(p){\mathrm d}p = nRT \int_{P_\mathrm{max}}^{P_\mathrm{min}} \frac{\mathrm{d}p}{P} = nRT \log(\frac{P_\mathrm{min}}{P_\mathrm{max}})$$.

Comparison of the two
The ratio is
 * $$\frac{w_\mathrm{adiabatic}}{w_\mathrm{isothermal}} = {\frac{7}{2} \left[ \left( \frac{P_\mathrm{min}}{P_\mathrm{max}}\right)^{2/7} - 1 \right]}/{\log(\frac{P_\mathrm{min}}{P_\mathrm{max}})}$$,

which seems to be always greater than one. There is an error in this somewhere still.

units version 1.80 with readline, units database in /usr/share/units.dat 2084 units, 71 prefixes, 32 nonlinear units (1/10)^(2/7): Definition: 0.51794747 log(1/10): Definition: -1 ln(1/10): Definition: -2.3025851 (1/10)^(2/7)+-1: Definition: -0.48205253 7/2((1/10)^(2/7)+-1): Definition: -7.2606195 7/2((1/10)^(2/7)+-1)/ln(1/10): Definition: 3.153247 7/2((1/100)^(2/7)+-1)/ln(1/100): Definition: 1.0386548 7/2((1/2)^(2/7)+-1)/ln(1/2): Definition: 28.104765 7/2(((1/2)^(2/7))+(-1))/ln(1/2): Definition: 28.104765 7/2(((1/2)^(2/7))+(-1)): Definition: -19.480739 ln(1/2): Definition: -0.69314718 ln(300): Definition: 5.7037825