Question

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

• A. $12.42\times {10}^{-21}$ J, $968$ m/s
• B. $8.78\times {10}^{-21}$ J, $684$ m/s
• C. $6.21\times {10}^{-21}$ J, $968$ m/s
• D. $12.42\times {10}^{-21}$ J, $684$ m/s

Answer 1

The correct option is D $12.42\times {10}^{-21}$ J, $684$ m/s

$\begin{array}{c}E=averageenergy=6.21\times {10}^{-21}j\\ V_{rms}=rmsspeed=484m/s\\ T_1=300K\end{array}$

As $E\times T$, if T = 600K, temperature is doubled so energy (E) will also be doubled.

$\begin{array}{c}\therefore E=2\times 6.21\times {10}^{-21}j=12.42\times {10}^{-21}j\\ Also{V}_{rms}=\sqrt{3K\frac{bT}{m}},V_{rms}^2\times T\\ \frac{V_{rms}^2\left(600K\right)}{v=V_{rms}^2\left(300K\right)}=\left(\frac{600}{300}\right)=2\Rightarrow V_{rms}^2\left(600K\right)=2\times {484}^2\\ \Rightarrow 684.7m/s\end{array}$