Question

An electric bulb of 250cc was sealed off at a pressure ${10^{ - 3}}mm$ of Hg and temperature ${27^ \circ }C$. The number of molecules present in the gas is
A. $8.178 \times {10^{15}}$
B. $6.023 \times {10^{23}}$
C. $8.021 \times {10^{23}}$
D. $6 \times {10^{22}}$

Answer 1

An electric bulb is sealed under a certain temperature and pressure of mercury. To find the number of molecules, the Equation of state is used. The pressure, volume and temperature of a gas can describe the number of moles contained in that gas. The number of moles multiplied with the Avogadro’s number will give the number of molecules in the given gas.

Formula Used:
The pressure $P$ is given by: $P = \rho hg$
where, $\rho$ is the density of the gas, $h$ is the height of the gas, $g$ is the acceleration due to gravity = $10m/{s^2}$
The Equation of state is given by: $PV = nRT$
where, $R$ is the gas constant, $P$ is the pressure, $V$ is the volume, $T$ is the temperature, $n$ is the number of moles.

Complete step by step answer:
The values given in the problem are;
Volume of electric bulb = $V = 250cc = 250 \times {10^{ - 6}}cubic.meter$
Temperature = $T = {27^ \circ }C = 300K$
Height of Hg when the electric bulb was sealed = ${10^{ - 3}}mm = {10^{ - 3}} \times {10^{ - 3}}m$
Density of mercury =${\rho _{Hg}} = 13600kg/{m^3}$
Then, the pressure $P$ can be calculated by using the formula
$P = \rho hg$
For mercury, the formula for pressure will be;

\Rightarrow P = 13600 \times {10^{ - 3}} \times {10^{ - 3}} \times 10 \\\ \Rightarrow P = 136 \times {10^{ - 3}} \\\ \therefore P = 0.136\,Pa$$ The relation between pressure$$P$$, volume $$V$$ and temperature $$T$$ for the given sample of a gaseous system is called the equation of state. For a given system $$n$$ moles of perfect gas, the equation of state is $$PV = nRT$$ where, $$R$$ is the gas constant. $$R = 8.3145J.mo{l^{ - 1}}{K^{ - 1}}$$. $$n = \dfrac{{PV}}{{RT}} \\\ \Rightarrow n = \dfrac{{0.136 \times 250 \times {{10}^{ - 6}}}}{{8.3145 \times 300}} \\\ \Rightarrow n = 0.013630 \times {10^{ - 6}} \\\ \therefore n = 1.363 \times {10^{ - 8}}$$ The number of molecules are given by $$N = A \times n$$ where, $$N$$ is Avogadro's number $$N = 6 \times {10^{23}}mo{l^{ - 1}}$$ $$N = 6 \times {10^{23}} \times 1.363 \times {10^{ - 8}} \\\ \therefore N = 8.178 \times {10^{15}}$$ Therefore, there are $$8.178 \times {10^{15}}$$ molecules present in the gas. **Hence, option A is the correct answer.** **Note:** The Equation of state given above is only applicable for ideal gases. For real gases, the Equation of state is $$\left( {P + \dfrac{{a{n^2}}}{{{V^2}}}} \right)\left( {V - nb} \right) = nRT$$ where, $$a$$and $$b$$are substance specific constants. The Equation of state differs for different types of processes. For example, in the isothermal process $$PV = {\text{constant}}$$. In isochoric process $$\dfrac{V}{T} = {\text{constant}}$$ and for isobaric process, $$\dfrac{P}{T} = {\text{constant}}$$.