Question

A cubical box is side $1$ meter contains helium gas(atomic weight $4$) at a pressure of $100N/m^2$. During an observation time of $1$ second, an atom travelling with the root-mean-square speed parallel to one of the edges of the cube, was found to make $500$ hits with a particular wall, without any collision with other atoms.
Take $R=\frac{25}{3}$ J/mol-K and $k=1.38\times {10}^{-23}$ J/K.
Evaluate the temperature of the gas.

• A. $T=360$K.
• B. $T=160$K.
• C. $T=260$K.
• D. $T=460$K.

Answer 1

The correct option is B $T=160$K.

The volume of the box $1m^3$

Pressure of the gas $=100N/m^2$

Time between two consecutive collisions with one wall $=\frac{1}{500}s$

Let $T$ be the temperature of the gas

The time of collision should be equal to $\frac{2l}{v_{rms}}$, where $l$ is the length of the cube.

So, $\frac{2l}{v_{rms}}=\frac{1}{500}$

$v_{rms}=1000m/s$

The RMS velocity of the particle is given by:

$v_{rms}=\sqrt{\frac{2RT}{M}}$

$\therefore \sqrt{\frac{2RT}{M}}=1000$

$T=\frac{(1000{)}^{2}M}{3R}=\frac{\left({10}^6\right)(4\times {10}^{-3})}{3\left(\frac{25}{3}\right)}=160K$