Question

At $27^{\circ }C$ temperature the kinetic energy of an ideal gas is $E_1.$ If the temperature is increased to $327^{\circ }C$ then the kinetic energy will be :

• A. $\frac{E_1}{\sqrt{2}}$
• B. $\sqrt{2}{E}_1$
• C. $2E_1$
• D. $\frac{E_1}{2}$

Answer 1

Answer: (C)

Let the mass of $1g$ molecule of a gas be $M$, and its mean square velocity be $v^2$, then kinetic energy of $1g$ molecule of gas is
$\frac{1}{2}M{v}^2=\frac{1}{2}M\left(\frac{3RT}{M}\right)=\frac{3}{2}RT$
There are $N$-molecules in $1g$ molecule of gas.
Thus, average kinetic energy of one molecule
$=\frac{(3/2)RT}{N}$
$=\frac{3}{2}\left(\frac{R}{N}\right)T$
where, $(R/N)=k$ is Boltzmann's constant.
$\therefore KE=\frac{3}{2}kT$.
Given,
$T_1=27^{\circ }C=273+27=300K$
$T_2=327^{\circ }C=273+327=600K$
$\therefore \frac{(KE{)}_2}{(KE{)}_1}=\frac{600}{300}=\frac{2}{1}$
$\Rightarrow (KE{)}_2=2(KE{)}_1$
$\Rightarrow E_2=2E_1$