Question

An ideal gas mixture inside a balloon expands according to the relation $P{V^{2/3}} = {\rm{ constant}}$. The temperature inside the balloon is
(A) increasing
(B) decreasing
(C) constant
(D) can’t be said

Answer 1

Hint: You need to use the concept of the ideal gas equation in order to solve the question. You need to substitute the expression of an ideal gas in the relation given in the question. Later you will obtain the expression that will help you to get the answer.

Complete step by step answer:
It is given that a gas is filled inside the balloon that is having characteristics similar to that of ideal gas. The gas inside the balloon expands that can be shown by the relation $P{V^{2/3}}$= constant.

As we know that the expression for the ideal gas equation is given as,
$PV = nRT$
Here, P is the pressure, V is the volume of the gas, n is the number of moles, R is the gas constant, and T is the temperature of the gas.

We can rewrite the above expression as,
$P = \dfrac{{nRT}}{V}$

Here it is given that the relation is $P{V^{2/3}}$= constant.

Substitute the value of ideal gas equation in the expression above,
$
P{V^{2/3}} = {\rm{constant}}\\
\implies \dfrac{{nRT}}{V} \times {V^{2/3}} = {\rm{constant}}\\
\implies T{V^{ - 1/3}} = {\rm{constant}}
$

Again rearranging the above expression, we will get,
$T = {\rm{constant}} \times {V^{1/3}}$

So from the above expression that we obtained, we can clearly see that the temperature of the gas is linearly proportional to the volume of the gas. So we can rewrite it as,
$T \propto {V^{1/3}}$

We can here say that, when the volume of the ideal gas inside the balloon decreases, the temperature also decreases. Whereas, when the volume of the ideal gas increases in the balloon, the temperature of the ideal gas inside the balloon also increases.

So, the correct answer is “Option A”.

Note:
You can make a mistake while substituting the expression of the ideal gas equation in the relation given in the question. You cannot solve the question directly without solving it. Otherwise, you can go wrong. As the volume is linearly proportional to the temperature, so it will increase as the volume will increase.