Isothermal and adiabatic expansion

Suppose that the temperature of an ideal gas is held constant by keeping the gas in thermal contact with a heat reservoir. If the gas is allowed to expand quasi-statically under these so called isothermal conditions then the ideal equation of state tells us that

$$p\,V = {\rm constant}.$$ (311)

This is usually called the isothermal gas law.

Suppose, now, that the gas is thermally isolated from its surroundings. If the gas is allowed to expand quasi-statically under these so called adiabatic conditions then it does work on its environment, and, hence, its internal energy is reduced, and its temperature changes. Let us work out the relationship between the pressure and volume of the gas during adiabatic expansion.

According to the first law of thermodynamics,

$${\mathchar'26\mskip-12mud}Q = \nu \,c_V \,dT +p\,dV = 0,$$ (312)

in an adiabatic process (in which no heat is absorbed). The ideal gas equation of state can be differentiated, yielding

$$p\,dV + V\,dp = \nu\, R \,dT.$$ (313)

The temperature increment $dT$ can be eliminated between the above two expressions to give

$$0 = \frac{c_V}{R} (p\,dV + V\,dp) + p \,dV = \left(\frac{c_V}{R} + 1\right) \,p\, dV +\frac{c_V}{R} \,V\,dp,$$ (314)

which reduces to

$$(c_V +R)\,p\,dV + c_V \,V\, dp = 0.$$ (315)

Dividing through by $c_V\, p\, V$ yields

$$\gamma\,\frac{dV}{V} + \frac{dp}{p}=0,$$ (316)

where

$$\gamma \equiv \frac{c_p}{c_V} = \frac{c_V + R}{c_V}.$$ (317)

It turns out that $c_V$ is a very slowly varying function of temperature in most gases. So, it is always a fairly good approximation to treat the ratio of specific heats $\gamma$ as a constant, at least over a limited temperature range. If $\gamma$ is constant then we can integrate Eq. (316) to give

$$\gamma \ln V + \ln p = {\rm constant},$$ (318)

or

$$p \,V^{\,\gamma} = {\rm constant}.$$ (319)

This is the famous adiabatic gas law. It is very easy to obtain similar relationships between $V$ and $T$ and $p$ and $T$ during adiabatic expansion or contraction. Since $p = \nu\, R\,T/V$, the above formula also implies that

$$T \,V^{\,\gamma-1} = {\rm constant},$$ (320)

and

$$p^{\,1-\gamma} \,T^{\,\gamma} = {\rm constant}.$$ (321)

Equations (319)-(321) are all completely equivalent.