Question

You might be familiar with the ideal gas equation from Chemistry, which relates the pressure and volume of an ideal gas with temperature as: $PV$ = $nRT$, where $nR$ is a constant. You are given a closed metal container of volume $V$, filled with a gas kept at a temperature $T$, and pressure $P$. If $\gamma$ is the coefficient of volume expansion for the gas, what is the pressure after the temperature is increased by $\Delta T$?

• A. $P^{\prime }$ = $P(1+\gamma \Delta T)$
• B. $P^{\prime }$ = $P(1-\gamma \Delta T)$
• C. $P^{\prime }$ = $P$
• D. $P^{\prime }$ = $\frac{P}{(1+\gamma \Delta T)}$

Answer 1

The correct option is A

$P^{\prime }$ = $P(1+\gamma \Delta T)$

Let's consider the second case first where the pressure is fixed due to the elasticity of the balloon, and the volume is allowed to increase. According to the ideal gas equation,

$PV$ = $nRT$

or, $V$ = $\frac{nR}{P}$ $T$.

At a new temperature $T^{\prime }$, the volume will be -

$V^{\prime }$ = $\frac{nR}{P}$ $T^{\prime }$. - - - - - - (1)

We have a different expression for $V^{\prime }$ using the coefficient $\gamma$, too, in the following way -

$V^{\prime }$ = $V(1+\gamma \Delta T)$

$\Rightarrow$ $\frac{nR{T}^{\prime }}{P}$ = $\frac{nRT}{P}(1+\gamma \Delta T)$

$\Rightarrow$ $T^{\prime }$ = $T$ + $T\gamma \Delta T$

$\Rightarrow$ $(T^{\prime }-T)$ = $T\gamma \Delta T$

$\Rightarrow$ $\Delta T$ = $T\gamma \Delta T$

$\Rightarrow$ $\gamma$ = $\frac{1}{T}$ - - - - - - (2)

This is an unexpected result, is not it? It means, the coefficient of volume expansion for an ideal gas is not a constant at all, but depends on the reciprocal of the temperature, at which we are measuring it.

But we are asked to find the new pressure when the volume is kept constant, and not the other way round. So, we need some kind of coefficient for "pressure” expansion - let's call it $\omega$.

Just like before, we can write ($P$, instead of $V$) -

$P^{\prime }$ = $P$ ($1+\omega \Delta T$)

$\Rightarrow$ $\frac{nR{T}^{\prime }}{V}$ = $\frac{nRT}{V}(1+\omega \Delta T)$

$\Rightarrow$ $T^{\prime }$ = $T+T\omega \Delta T$

$\Rightarrow$ $\omega$ = $\frac{1}{T}$ .............(3)

Wait! There was no reason to believe that the coefficient for pressure expansion would be the same as volume expansion, but it turns out (compare equation (2) and (3)), at a temperature $T$, both $\gamma$ and $\omega$ are equal to $\frac{1}{T}$. This is wonderful and makes the problem so much simpler.

Thus, we can write the new pressure $P^{\prime }$ as just -

$P^{\prime }$ = $P(1+\gamma \Delta T)$,

where we have used $\gamma$ instead of $\omega$.