Question

The number of molecules in a volume of $4c{m}^3$a perfect monatomic gas at some temperature$T$ and pressure of$2cm$ of mercury is close to? (Given, that the mean kinetic energy of a molecule (at $T$)is$4\times {10}^{-14}erg$, $g=980cm{s}^{-2}$the density of mercury = $13.6gc{m}^{-3}$)

• A. $4·0\times {10}^{18}$
• B. $4·0\times {10}^6$
• C. $5·8\times {10}^{16}$
• D. $5·8\times {10}^{18}$

Answer 1

The correct option is A

$4·0\times {10}^{18}$

Step 1. Given data:

1. Volume of a monoatomic gas$V=4c{m}^3$
2. Fluid pressure$P=2cmofmercury$
3. Height of fluid column$h=2cm$
4. The mean kinetic energy of a molecule (at $T$)$E=4\times {10}^{-14}erg$
5. Gravitational acceleration$g=980cm{s}^{-2}$
6. The density of mercury$\rho =13·6gc{m}^{-3}$

Step 2. Formula used:

The ideal gas equation for the perfectly monoatomic gas is given by,

As we know from Boyle's law,

$$\begin{gathered} P\propto \frac{1}{V} \\ PV=Cons\tan t\left(T=cons\tan t\right) \end{gathered}$$

$$\begin{gathered} PV=NkT(k=Boltzmancons\tan t) \\ E=\frac{3}{2}kT \end{gathered}$$

Step 3. Calculating the number of molecules:

The number of molecules in the perfectly monoatomic gas can be calculated as,

$\begin{array}{rcl}N& =& \frac{PV}{kT}\\ & =& \frac{3PV}{2E}(as,kT=\frac{2E}{3})\\ & =& \frac{3\rho gh}{2E}(as,P=\rho gh)\end{array}$

$\begin{array}{rcl}N& =& \frac{3\times 13·6\times 980\times 2\times 4}{2\times 4\times 10-14}\\ & =& 3·998\times {10}^{18}\\ & \simeq & 4\times {10}^{18}\end{array}$

Thus, the number of molecules is$\begin{array}{rcl}& & 4\times {10}^{18}\end{array}$.

Hence, option A is the correct answer.