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 (assuming ideal gas behaviour)

• 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

Answer: (D)

12.42 $\times 10^{-21}$ J, 684 m/s

Answer 2

The average translational KE = $\frac{3}{2}kT$ which is directly proportional to T, while rms speed of molecules is given by

$\, \, \, \, \, \, \, \, \, v_{rms}=\sqrt{\frac{3RT}{M}} \, i.e.,v_{rms} \, = \sqrt T$
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 =$2 \times 6.21 \, \times 10^{-21} J$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =12.42 \times 10^{-21} J$
and $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, v_{rms}=(1.414)(484)m/s$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =684 m/s$