Question

One mole of oxygen at ${27}^o$C and at one atmospheric pressure is enclosed in a vessel. Assuming the molecules to be moving with $V_{rms}$. Find the number of collisions per second which the molecules make with $1$ $m^2$ area of the vessel wall.

• A. $4.4\times {10}^{27}$.
• B. $1.97\times {10}^{27}$.
• C. $5.4\times {10}^{27}$.
• D. $6.4\times {10}^{27}$.

Answer 1

Answer: (B)

$1.97\times {10}^{27}$.

Solution:

$F=P\times A={10}^5\times 1={10}^{5}N$

$F=\Delta P\Delta t$ $\to$ $\delta p=F\times \delta t={10}^5\times 1={10}^5$…..(i)

Now, momentum change per second

$\left(\Delta p\right)=n\times 2mv$…..(ii)

Where n is the number of collisions per second per

From (i) and (ii)

$n\times 2mv={10}^5$

$\therefore n=\frac{{10}^5}{2mv}$

Root mean square velocity

$v=\sqrt{\frac{3RT}{M}}$ = $\sqrt{\frac{3\times 8.314\times 300}{32/1000}}$ =483.4m/s

According to mole concept $6.023\times {10}^{23}$g molecules will

have mass 32g

will have mass $\frac{32}{6.023\times {10}^{2}3}$g

$\therefore n=\frac{{10}^5\times 6.023\times {10}^{2}3}{2\times 32\times 483.4}=1.97\times {10}^{27}$