Question

One mole of oxygen at ${27}^o$C and at one atmospheric pressure is enclosed in a vessel. The vessel is now thermally insulated and moved with a constant speed $V_0$. It is then suddenly stopped. The process results in a rise of the temperature of the gas by $1^o$C. Calculate the speed $V_0$.

• A. $65.6$ m/s.
• B. $35.6$ m/s.
• C. $55.6$ m/s.
• D. $75.6$ m/s.

Answer 1

The correct option is B $35.6$ m/s.

$\text{Solution}:$

The kinetic energy of motion of molecules will be converted into heat energy.

K.E. of $1gm$ mole of oxygen=$\frac{1}{2}M{V}_0^2...\left(i\right)$

Where $v_0$ is the velocity with which the vessel was moving.

The heat gained by 1 gm mole of molecules at constat volume for $1^{o}C$ rise in temperature

$=n{C}_v\Delta T=1\times C_v\times 1=C_v.....\left(ii\right)$

From (i) and (ii)

$\frac{1}{2}M{V}_0^2=C_v$

But, $C_v=\frac{R}{\gamma -1}\frac{1}{2}M{V}_0^2=\frac{R}{\gamma -1}$

$\therefore v_0=\sqrt{\frac{2R}{M\gamma -1}}=\sqrt{\frac{2\times 8.314}{\frac{32}{100}\times (1.41-1)}}=35.6m{s}^{-1}$

$\text{Hence B is the correct option}$