Isothermal and adiabatic

Problem

From the same point on the $p\text{-}V$ plane draw an isotherm and a reversible adiabat. Explain in five lines why the adiabat drops more steeply.

Solution

Slope of the isotherm. For an ideal gas at constant temperature $pV=\text{const}$. Differentiating: $p\,\dd V + V\,\dd p = 0$, so the slope in the $p\text{-}V$ plane is

$\left(\frac{\dd p}{\dd V}\right)_{T} = -\frac{p}{V}$

Slope of the adiabat. In a reversible adiabatic process $pV^{\gamma}=\text{const}$. Differentiating: $\gamma p V^{\gamma-1}\,\dd V + V^{\gamma}\,\dd p = 0$, so

$\left(\frac{\dd p}{\dd V}\right)_{\text{ad}} = -\gamma\,\frac{p}{V}$

Comparison. Since $\gamma = c_p/c_V > 1$, at the same point (same $p$ and $V$) the adiabat has a greater slope in magnitude: it drops more steeply.

Physical interpretation. In the adiabatic expansion the gas receives no heat ($Q=0$): the energy to do work comes from the internal energy, so the gas cools and the pressure falls more. In the isothermal process, instead, the gas absorbs heat that compensates for the work, keeps $T$ constant and the pressure falls less.

$\ev{\text{adiabat steeper because } \gamma>1}$

Links

Topics: Thermodynamics Concepts: Thermodynamic processes Methods: P-V and ln P - ln V diagram Objects: Ideal gas