A) 516m/s
B) 450 m/s
C) 310 m/s
D) 746 m/s
Correct Answer: A
Solution :
The root mean square velocity of the gas is given by \[{{v}_{rms}}=\sqrt{\frac{3RT}{M}}\] where, R is gas constant, T is absolute temperature and m is the molecular weight of the gas. \[{{T}_{1}}=27{}^\circ C=273+27=300K\] \[{{T}_{2}}=227{}^\circ C=273+227=500K\] \[\frac{{{v}_{1}}}{{{v}_{2}}}=\sqrt{\frac{300}{500}}=\sqrt{\frac{3}{5}}\] Given \[{{v}_{1}}=400\,m/s,{{v}_{2}}={{v}_{s}}\] \[{{v}_{s}}=\sqrt{\frac{5}{3}}\times 400\] \[=129\times 400\] \[=516.39\text{ }m/s=516\text{ }m/s\] Note If the absolute temperature of the gas become zero then the motion of molecules will cease.You need to login to perform this action.
You will be redirected in
3 sec