$$N$$ molecules each of mass $$m$$ of a gas $$A$$ and $$2N$$ molecules each of mass $$2\,m$$ of gas $$B$$ are contained in the same vessel which is maintained at temperature $$T.$$ The mean square velocity of molecules of $$B$$ type is $${v^2}$$ and the mean square rectangular component of the velocity of $$A$$ type is denoted by $${\omega ^2}.$$ Then $$\frac{{{\omega ^2}}}{{{v^2}}}$$ is
A.
$$2$$
B.
$$1$$
C.
$$\frac{1}{3}$$
D.
$$\frac{2}{3}$$
Answer :
$$\frac{2}{3}$$
Solution :
Mean kinetic energy of the two types of molecules should be equal. The mean From question, square velocity of $$A$$ type molecules $$ = {\omega ^2} + {\omega ^2} + {\omega ^2} = 3{\omega ^2}$$
Therefore, $$\frac{1}{2}m\left( {3{\omega ^2}} \right) = \frac{1}{2}\left( {2m} \right){v^2}$$
This gives $$\frac{{{\omega ^2}}}{{{v^2}}} = \frac{2}{3}$$
Releted MCQ Question on Heat and Thermodynamics >> Kinetic Theory of Gases
The average translational kinetic energy of $${O_2}$$ (relative molar mass 32) molecules at a particular temperature is $$0.048\,eV.$$ The translational kinetic energy of $${N_2}$$ (relative molar mass 28) molecules in $$eV$$ at the same temperature is
A vessel contains 1 mole of $${O_2}$$ gas (relative molar mass 32) at a temperature $$T.$$ The pressure of the gas is $$P.$$ An identical vessel containing one mole of $$He$$ gas (relative molar mass 4) at a temperature $$2\,T$$ has a pressure of