Question 5.6: For an ideal gas, show that at any point on a Clapeyron (p, ...

Principles of Thermodynamics [EXP-7769]

For an ideal gas, show that at any point on a Clapeyron (p, V) diagram, the absolute value of the slope is greater for an adiabatic process (A) than an isothermal process (I).

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An isothermal process (I) in a Clapeyron diagram is characterised by,

$p V \equiv C_I = const$ where $C_I = N R T$ .

Thus, for an isothermal process,

$p = \frac{C_I}{V}$ and $\frac{dp}{dV} = – \frac{C_I}{V^2}= – \frac{NR T}{V} \frac{1}{V} = -\frac{p}{V}$ .

An adiabatic process (A) in a Clapeyron diagram is characterised by,

$p V^γ \equiv C_A = const$ where $C_A = p^{1−γ} (NR T)^γ$ .

Thus, for an adiabatic process,

$p = \frac{C_A}{V^γ}$ and $\frac{dp}{dV} = – \frac{γ C_A}{V^{γ+1}} = – γ \Bigl(\frac{NR T}{pV}\Bigr) ^γ \frac{p}{V} = – γ \frac{p}{V}$ .

The slopes of both processes are negative in the Clapeyron diagram. Since γ > 1, the absolute value of the slope of the adiabatic process is greater than that of the isothermal process.

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