Expansion coefficient for an ideal gas at constant pressure is $α = 1/Τ$. Explain this relation

Question

How the coefficient of expansion for an ideal gas is given by $α=1/Τ$, (at constant pressure). How could it be inversely proportional to it. Doesn't gas expand more on higher temperature, but this relation is different from what i can imagine. A little help to this would be really great. Thanks in advance

Answer 1

The thermal coefficient of expansion of a substance is the proportional increase in volume for a $1$ Kelvin rise in temperature.

For an ideal gas we know that

$PV = nRT$

so if pressure $P$ is constant we can express volume $V$ as a function of temperature $T$:

$\displaystyle V(T) = \frac {nR} P T$

So near $T=300$, for example, we have

$\displaystyle V(300) = 300 \frac {nR} P \\ \displaystyle V(301) = 301 \frac {nR} P \\ \displaystyle \Rightarrow \Delta V = V(301) - V(300) = \frac {nR} P$

In other words, the absolute increase in volume per $1$ Kelvin rise in temperature is $\Delta V = \frac {nR} P$ - and this does not depend on the temperature. But to find the proportional increase in volume we need to divide $\Delta V$ by $V(300)$:

$\displaystyle \alpha(300) = \frac {\Delta V}{V(300)} = \left( \frac {nR} {P} \right) \left( \frac {P} {300 nR}\right) = \frac 1 {300}$

Do the same calculation for any temperature and you will see that the proportional increase in volume is always $\frac 1 T$. In other words

$\displaystyle \alpha(T) = \frac 1 T$

One way to see this intuitively is to realise that if the absolute change in volume per $1$ Kelvin is constant, and volume increases as temperature rises, then the proportional change in volume per $1$ Kelvin must decrease as temperature rises.