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