Question

The mass of hydrogen molecules is $3.32\times {10}^{-24}$ gm. If ${10}^{23}{H}_2$ molecules strike $2$ sq. cm are per second with velocity of ${10}^5$ cm/sec at an angle of ${45}^{\circ }$ to the normal to wall, then the exerted pressure will be

• A. $2.35N/m^2$
• B. $23.5N/m^2$
• C. $235N/m^2$
• D. $2350N/m^2$

Answer 1

The correct option is D $2350N/m^2$

$\begin{array}{c}\Delta P=\left(2mv\cos {45}^o\right)\times {10}^{23}\\ =2\times 3.32\times {10}^{-24}\times {10}^{-3}\times \frac{10}{100}\times \frac{1}{\sqrt{2}}\times 10\\ \Rightarrow F=\frac{2\times 3.32}{\sqrt{2}}\times {10}^{-1}\\ \therefore pressure=\frac{2\times 3.32\times {10}^{-1}}{\sqrt{2}\times 2\times {10}^{-4}}\\ =\frac{2\times 3.32}{2\sqrt{2}}\times {10}^3\\ =\frac{3.32}{\sqrt{2}}\times {10}^{3}N/m^2\\ \approx 2350N/m^2\\ Hence,\\ optionDiscorrectanswer.\end{array}$