Heat and Thermodynamics 4 Question 26
28. 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 (assuming ideal gas behaviour)
(1997, 1M)
(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) $12.42 \times 10^{-21} J, 684 m / s$
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Answer:
Correct Answer: 28. (d)
Solution:
- The average translational $KE=\frac{3}{2} k T$ which is directly proportional to $T$, while rms speed of molecules is given by
$$ v _{rms}=\sqrt{\frac{3 R T}{M}} \text { i.e. } v _{rms} \propto \sqrt{T} $$
When temperature of gas is increased from $300 K$ to $600 K$ (i.e. 2 times), the average translational $KE$ will increase to 2 times and rms speed to $\sqrt{2}$ or 1.414 times.
$\therefore$ Average translational KE $=2 \times 6.21 \times 10^{-21} J$
and
$$ =12.42 \times 10^{-21} J $$
$$ \begin{aligned} v _{rms} & =(1.414)(484) m / s \\ & \approx 684 m / s \end{aligned} $$
