Question

The average translational kinetic energy and the root mean square speed of the molecules in a sample of oxygen gas at $300\text{K}$ are $6.21\times {10}^{-21}\text{J}$ and $484\text{m/s}$ respectively. The corresponding values at $600\text{K}$ are nearly (assuming ideal gas behaviour)

• A. $2.42\times {10}^{-20}\text{J}$ and $968\text{m/s}$
• B. $8.78\times {10}^{-20}\text{J}$ and $684\text{m/s}$
• C. $6.21\times {10}^{-20}\text{J}$ and $968\text{m/s}$
• D. $1.24\times {10}^{-20}\text{J}$ and $684\text{m/s}$

Answer 1

The correct option is D $1.24\times {10}^{-20}\text{J}$ and $684\text{m/s}$

Given:
Initial temperature, $T_1=300\text{K}$
Initial average translational kinetic energy, $K{E}_1=6.21\times {10}^{-21}\text{J}$
Initial root mean square speed, $v_1=484\text{m/s}$
Final temperature, $T_2=600\text{K}$
To find:
Final average translational kinetic energy, $K{E}_2=?$
Final root mean square speed, $v_2=?$

We know that, average translational kinetic energy is given by,
$KE=\frac{nRT}{2}$
where,
$n\to$ number of moles
$R\to$ universal gas constant
$T\to$ temperature

So, $\frac{K{E}_1}{K{E}_2}=\frac{T_1}{T_2}$
[ $n$ remain same because gas is same.]
$\Rightarrow K{E}_2=\frac{K{E}_1\times T_2}{T_1}$
$\Rightarrow K{E}_2=\frac{6.21\times {10}^{-21}\times 600}{300}$
$\Rightarrow K{E}_2=1.24\times {10}^{-20}\text{J}$

Now,
Root mean square speed is given by,
$v=\sqrt{\frac{3RT}{M}}$
So, $\frac{v_1}{v_2}=\sqrt{\frac{T_1}{T_2}}$
[$M$ remains same]
$\Rightarrow v_2=v_1\times \sqrt{\frac{T_2}{T_1}}$
$\Rightarrow v_2=484\times \sqrt{\frac{600}{300}}$
$\Rightarrow v_2\approx 684\text{m/s}$