The molecules of a given mass of a gas have r.m.s. velocity of 200 ms^ -1 at 27^ ^ C and 1.0 × 10^ 5 Nm^ -2 pressure. When the temperature and pressure of the gas are respectively, 127^ ^ C and 0.05 × 10^ 5 Nm^ -2 , the

Correct answer: B. 400 3 Here v_ 1 =200 m / s ; temperature T_ 1 =27^ ^ C=27+273=300 k temperature T_ 2 =127^ ^ C=127+273=400 kV=? R.M.S. Velocity, V T array l (v)/(200)= (400)/(300) v= 200 × 2 3 m/ s v= 400 3 m/ s array

BITSAT 2016

Exam: BITSAT
Year: 2016
Subject: Physics
Chapter: Kinetic Theory of Gases

BITSAT 2016 · Physics · Kinetic Theory of Gases

The molecules of a given mass of a gas have r.m.s. velocity of 200...

The molecules of a given mass of a gas have r.m.s. velocity of

200 \mathrm{ ms}^{-1}

27^{{}^{\circ}} \mathrm{C}

and

1.0 \times 10^{5}\mathrm{Nm}^{-2}

pressure. When the temperature and pressure of the gas are respectively,

127^{{}^{\circ}} \mathrm{C}

and

0.05 \times 10^{5} \mathrm{Nm}^{-2},

the r.m.s. velocity of its molecules in

\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm\mathrm{ms}^{-1}

is :

Options

A $100 \sqrt{2}$
B $\frac{400}{\sqrt{3}}$
C $\frac{100 \sqrt{2}}{3}$
D $\frac{100}{3}$

Correct answer

B. $\frac{400}{\sqrt{3}}$

Step-by-step solution

Here

v_{1}=200 \mathrm{ m} / \mathrm{s}

; temperature

\mathrm{T}_{1}=27^{{}^{\circ}} \mathrm{C}=27+273=300 \mathrm{k}

temperature

\mathrm{T}_{2}=127^{{}^{\circ}} \mathrm{C}=127+273=400 \mathrm{k}\mathrm{V}=?

R.M.S. Velocity,

\mathrm{V} \propto \sqrt{\mathrm{T}}

\begin{array}{l} \Rightarrow \quad\frac{v}{200}=\sqrt{\frac{400}{300}}\\ \Rightarrow \quad v=\frac{200 \times 2}{\sqrt{3}} \mathrm{ m}/ \mathrm{s}\\ \Rightarrow \quad \mathrm{v}=\frac{400}{\sqrt{3}} \mathrm{ m}/ \mathrm{s} \end{array}

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