Question

One gram mole of oxygen at 27$^{\circ}$C and one atmospheric
pressure is enclosed in a vessel, (a) Assuming the molecules
to be moving with v$_{rms}$, find the number of collisions per
second which the molecules make with one square metre area
of the vessel wall, (b) The vessel is next 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$^{\circ}$C. Calculate the speed v$_0$.

Answer 1

(a) $v_{rms} =\sqrt{\frac{3Rt}{M}} =\sqrt{\frac{3 \times 8.31 \times 300 }{32 \times 10^{-3}}}$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, =483.4 m/s$
Given, p$_0 = 1.01 \times 10^5N/m^2$= Force per unit area.
Let n molecules of oxygen strike the wall per second per
m$^2$ and recoil with same speed. Change in momentum is
(2 nmv$_{rms}$).
The change in momentum per unit time is the force.
Hence, $p_0 =2nmv_{rms}$
$\therefore \, \, \, \, \, \, \, n=\frac{p_0}{2mc_{rms}} = \frac{1.01 \times 10^5}{2\bigg[\frac{32}{6.02 \times 10^{26}}\bigg](483.4)}$
$ \, \, \, \, \, \, \, \, \, \, \, \, =1.96 \times 1^{27} / s$
(b) $\frac{1}{2} (m_{gas})v_0^2 =nC_v \Delta T$
$\therefore \, \, \, \, \, \, v_0 =\sqrt{\frac{2nC_v \Delta t}{m_{gas}}} =\sqrt{\frac{(2)(n)\bigg(\frac{5}{2} \times 8.31\bigg)(1)}{(n)(32 \times 10^{-3})}}$
$ =36 m/s$