Q. In a certain region of space there are only 5 molecules per $cm^{2}$ on an average. The temperature there is 3 K. The pressure of this dilute gas is ($k = 1.38 \times 10^{- 23}\ JK^{- 1})$
A) $20.7 \times 10^{- 17}\ Nm^{- 1}$
B) $15.3 \times 10^{- 13}\ Nm^{- 1}$
C) $2.3 \times 10^{- 10}\ Nm^{- 1}$
D) $5.3 \times 10^{- 5}\ Nm^{- 1}$
Explanation
(a) From ideal gas equation
$pV = nkT$
$p = \frac{n\ }{V}kT$
Here, $\frac{n}{V} = \frac{5}{cm^{3}} = 5 \times 10^{6}/m^{3}$
$\therefore\ \ \ \ \ \ p = (5 \times \frac{10^{6}}{m^{3})(1.38 \times \frac{10^{- 23}}{JK^{- 1}) \times 3K}}$
$p = 20.7 \times 10^{- 17}\ Nm^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.