Chapter 13, Problem 53A

Interpretation Introduction

Interpretation: This is to be explained that on increasing the temperature the average kinetic energy of the gas molecules increases.

Concept Introduction: - When the temperature of the gas is raised, the molecules present in the gases gets more energy. So, their speed is increased. They bombard with each other as well as on the walls of the container. When they bombard on the walls, they exert pressure.

Answer to Problem 53A

The pressure produced due to collision of gas molecules should increase the temperature.

Explanation of Solution

On increasing the temperature, collision of molecules increases due to increase in speed of gas molecules and pressure also increases.

So, when temperature increases, kinetic energy of molecules increases and pressure increases.

The equation is

$\mathrm{PV}=\frac{1}{3}mn{v}^2$

Where,

P = pressure exerted by gas.

V = volume of gas.

m = mass of each molecule

n = total number of molecules

v = root mean square velocity.

According to the kinetic theory there is a direct relation between absolute temperature and average kinetic energy of the molecules.

  $\begin{array}{l}Kineticenergy=\frac{1}{2}mn{v}^2\propto T\\ \frac{1}{2}mn{v}^2=kT(k=constant)\end{array}$

  $\begin{array}{l}\frac{3}{2}\times \frac{1}{3}mn{v}^2=kT\\ \frac{1}{3}mn{v}^2=\frac{2}{3}kT\end{array}$

But,

  $\frac{1}{3}mn{v}^2=PV$ {Kinetic gas equation}

  $\begin{array}{l}PV=\frac{2}{3}kT\\ \frac{PV}{T}=\frac{2}{3}k\end{array}$

For 1 mole of gas constant represented by R.

  $\begin{array}{l}\frac{PV}{T}=R\\ PV=RT\end{array}$

This is general gas equation.

Thus, it is proved that kinetic energy increases on increasing the temperature.