19.7.4 Proof of Equation 19.34

Now let’s see where Equation 19.34 comes from. Consider an adiabatic process in which an infinitesimal amount of work dW done on a gas causes an infinitesimal change in the thermal energy. For an adiabatic process, with $d Q = 0$ , the first law of thermodynamics is

d E_{th} = d W

(19.36)

We can use Equation 19.31, which is valid for any gas process, to write $d E_{th} =$ $n C_{V} d T$ . Earlier in the chapter we found that the work done during a small volume change is $d W = - p d V$ . With these substitutions, Equation 19.36 becomes

n C_{V} d T = - p d V

(19.37)

The ideal-gas law can now be used to write $p = n R T / V$ . The $n$ cancels, and the $C_{V}$ can be moved to the other side of the equation to give

\frac{d T}{T} = - \frac{R}{C_{V}} \frac{d V}{V}

(19.38)

We’re going to integrate Equation 19.38, but anticipating the need for $γ = C_{P} / C_{V}$ we can first use the fact that $C_{P} = C_{V} + R$ to write

\frac{R}{C_{V}} = \frac{C_{P} - C_{V}}{C_{V}} = \frac{C_{P}}{C_{V}} - 1 = γ - 1

(19.39)

Now we integrate Equation 19.38 from the initial state i to the final state f:

\int_{T_{i}}^{T_{f}} \frac{d T}{T} = - (γ - 1) \int_{V_{i}}^{V_{f}} \frac{d V}{V}

(19.40)

Carrying out the integration gives

ln (\frac{T_{f}}{T_{i}}) = ln {(\frac{V_{i}}{V_{f}})}^{γ - 1}

(19.41)

where we used the logarithm properties $log a - log b = log (a / b)$ and $c log a = log (a^{c})$ .

Taking the exponential of both sides now gives

\begin{array}{l} (\frac{T_{f}}{T_{i}}) = {(\frac{V_{i}}{V_{f}})}^{γ - 1} \\ T_{f} V_{f}^{γ - 1} = T_{i} V_{i}^{γ - 1} \end{array}

(19.42)

This was Equation 19.35. Writing $T = p V / n R$ and canceling $1 / n R$ from both sides of the equation give Equation 19.34:

p_{f} V_{f}^{γ} = p_{i} V_{i}^{γ}

(19.43)

This was a lengthy derivation, but it is good practice at seeing how the ideal-gas law and the first law of thermodynamics can work together to yield results of great importance.