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Chapter 18: Problem 50

Prove that the slope of an adiabat at a given point in a \(p V\) diagram is \(\gamma\) times the slope of the isotherm passing through the same point.

Short Answer

Expert verified
Yes, the slope of an adiabat at any point in a pV diagram is indeed \(\gamma\) times the slope of the isotherm passing through that same point.

Step by step solution

01

Defining terms

Assuming we have a monoatomic ideal gas, the equation for the adiabatic process is \(p V^\gamma = K1\) (where \(K1\) is a constant and \(\gamma = C_p/C_v\), the ratio of the specific heats at constant pressure to volume), and the equation for an isotherm process is \(p V = K2 - constant\). We are asked to prove that the slope of the adiabat at any point on the P-V diagram is equal to \(\gamma\) times the slope of the isotherm.
02

Differential of adiabat

Firstly, let's differentiate both sides of the adiabat equation \(p V^\gamma = K1\). Using the product rule and the chain rule, this gives us \(V^\gamma dp + \gamma p V^{\gamma -1} dV = 0\). Solving for \(dp/dV\) we find that the slope of adiabat is \( \frac{dp}{dV} = - \frac{\gamma p}{V} \).
03

Differential of isotherm

Next, differentiate both sides of the isotherm equation \(pV = K2\). This gives \(p dV + V dp = 0\). Solving for \(dp/dV\), we find that the slope of isotherm is \(\frac{dp}{dV} = - \frac{p}{V}\).
04

Comparing the slopes

Looking at both slopes, it is clear that the slope of the adiabat is \(\gamma\) times the slope of the isotherm, since \(- \frac{\gamma p}{V} = \gamma (- \frac{p}{V})\). Thus, the assertion is proven.

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