Problem — Isotherm on a logarithmic scale

Problem

An ideal gas at constant $T$ satisfies $pV = nRT$. On the plane $\log p$ vs $\log V$, the graph is a straight line. Find the slope and intercept in terms of $n$ and $T$.

Solution

At constant temperature the product $nRT$ is a constant. I find the pressure: $p = \frac{nRT}{V}$ I take the logarithm of both sides and use $\log(a/b) = \log a - \log b$: $\log p = \log(nRT) - \log V$ This is the equation of a straight line $y = m\,x + q$ with $y = \log p$ and $x = \log V$: $\log p = (-1)\cdot \log V + \log(nRT)$ $\ev{\text{slope } m = -1, \quad \text{intercept } q = \log(nRT)}$ The slope $-1$ is universal (it does not depend on the gas), while the intercept rises as $n$ or $T$ increase.

Links

Topics: Kinetic theory of gases Concepts: Ideal gas law Skills: Use of logarithmic scales · Reading graphs