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Chapter 18: Problem 44

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
The differential form of both, the equation for an adiabatic process and the equation of state for an ideal gas, shows that the slope of an adiabat at any given point in a pV diagram is indeed \(\gamma\) times the slope of the isotherm passing through the same point.

Step by step solution

01

Define the Equations

The adiabatic process equation can be written as \( pV^\gamma = C \), where \(C\) is a constant, \(\gamma = C_p/C_v\), and \(p\) and \(V\) are the pressure and volume, respectively. The equation of state for an ideal gas is \(pV = nRT\), where \(n\) is the amount of substance, \(R\) is the ideal, or universal, gas constant, and \(T\) is the absolute temperature.
02

Differentiate the Equations

Next, differentiate both sides of the adiabatic process equation with respect to volume, that will give us the slope of an adiabat. Likewise, differentiate both sides of the equation of state for an ideal gas, which will result in the slope of an isotherm. This yields \( \frac{dp}{dV} = -\frac{\gamma p}{V} \) for the adiabat and \( \frac{dp}{dV} = -\frac{p}{V} \) for the isotherm.
03

Compare the Slopes

\(\frac{dp}{dV} = -\frac{\gamma p}{V}\) for the adiabat and \( \frac{dp}{dV} = -\frac{p}{V} \) for the isotherm. Notice that the slope for the adiabat is exactly \(\gamma\) times the slope of the isotherm, at any given point as both equations are showing slopes at any given point in a \( pV \) diagram. This proves the original assertion.

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Most popular questions from this chapter

Are the initial and final equilibrium states of an irreversible process describable by points in a \(p V\) diagram? Explain.

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