Kinetic Theory - Result Question 13
13. The molecules of a given mass of a gas have r.m.s. velocity of $200 ms^{-1}$ at $27^{\circ} C$ and $1.0 \times 10^{5}$ $Nm^{-2}$ pressure. When the temperature and pressure of the gas are respectively, $127^{\circ} C$ and $0.05 \times 10^{5} Nm^{-2}$, the r.m.s. velocity of its molecules in $ms^{-1}$ is :
(a) $100 \sqrt{2}$
(b) $\frac{400}{\sqrt{3}}$
(c) $\frac{100 \sqrt{2}}{3}$
(d) $\frac{100}{3}$
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Answer:
Correct Answer: 13. (b)
Solution:
- (b) Here $v_1=200 m / s$;
temperature $T_1=27^{\circ} C=27+273=300 k$
temperature $T_2=127^{\circ} C=127+273=400{k, v_2}=$ ?
R.M.S. Velocity, $v \propto \sqrt{T}$
$\Rightarrow \frac{v_2}{200}=\sqrt{\frac{400}{300}}$
$\Rightarrow v_2=\frac{200 \times 2}{\sqrt{3}} m / s \Rightarrow v_2=\frac{400}{\sqrt{3}} m / s$
