Question

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

Answer 1

The Correct Option is A

Velocity and Temperature Relationship 

The relationship between velocity (\( v \)) and temperature (\( T \)) is given by:

\(v \propto \sqrt{T}\)

Now, let's solve for the velocity in the given scenario:

\(\frac{v}{200} = \sqrt{\frac{400}{300}} \Rightarrow v = \frac{200 \times 2}{\sqrt{3}} \, \text{m/s}\)

Simplifying this, we get:

\(v = \frac{400}{\sqrt{3}} \, \text{m/s}\)

Conclusion:

The velocity is \( \frac{400}{\sqrt{3}} \) meters per second.