Question #d1389

Jun 1, 2017

For ideal gas on a pV diagram, an adiabatic curve is always steeper than an isotherm if they intersect.

Explanation:

The ideal gas equation: $p V = n R T$

• $p$ is pressure
• $V$ is volume
• $n$ is number of moles
• $R$ is ideal gas constant
• $T$ is the absolute temperature

For an isothermal process, the right hand side of the ideal gas equation does not change. Which means

$p V = \text{constant}$

The slope at any point is given by

$\frac{\mathrm{dp}}{\mathrm{dV}} = - \frac{p}{V}$

using implicit differentiation or any other methods.

For adiabatic processes, the path would follow a curve of the following form:

$p {V}^{\gamma} = \text{constant}$

$\gamma$ in the above equation is treated as a constant and is meant to be the ratio of heat capacities at constant pressure, ${c}_{p}$, to heat capacity at constant volume, ${c}_{V}$. This can be derived using the first Law of Thermodynamics.

Differetiating to find the slope

$\frac{\mathrm{dp}}{\mathrm{dV}} = - \gamma \frac{p}{V}$

This looks very similar to the expression for the slope of isotherm, except that gamma is always > 1 and therefore the slope would be steeper.