Question

A cylinder of $\text{10 L}$capacity at $\text{300 K}$containing the $\text{He}$gas is used to fill balloons. The cylinder recorded a pressure of$\text{1}{{\text{0}}^{\text{-2 }}}\text{bar}$. The number of $\text{He}$atoms present in the cylinder is:
A) $4.82\times {{10}^{21}}$
B) $2.41\times {{10}^{23}}$
C) $2.41\times {{10}^{21}}$
D) $4.82\times {{10}^{23}}$

Answer 1

The question can be solved using the concept of ideal gas equation $\text{PV=nRT}$.The ideal gas equation relates the pressure , volume , temperature, and several moles of gas with each other. The number of particles present in the gas is found by the relation of Avogadro's number $\text{6}\text{.023 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{23}}}\text{mo}{{\text{l}}^{\text{-1}}}$ with the number of moles. Which is $\text{n=}\dfrac{\text{Number of partilcles}}{\text{(}{{\text{N}}_{\text{A}}}\text{)}}$

Complete step by step solution:
We are given the data as:
Pressure on the cylinder, $\text{P=1}{{\text{0}}^{\text{-2}}}\text{bar}$
The capacity of the cylinder,$\text{V=10L}$
Temperature,$\text{T=300K}$
We have to find the number of $\text{He}$atoms present in the cylinder.
We know the ideal gas equation as:
$\text{PV=nRT}$
Where P is the pressure of the gas, V is the volume of the gas, n stands for the amount of gas measured in terms of moles, R is the gas constant and T stands for the absolute temperature in kelvin.
Let’s first rearrange the equation concerning the number of moles (n)
$\text{n=}\dfrac{\text{PV}}{\text{RT}}$
Now substitute the values from the given data. We get,
$\text{n=}\dfrac{\text{(1}{{\text{0}}^{\text{-2}}}\text{ bar )(10 L)}}{\text{(8}\text{.314}\times \text{1}{{\text{0}}^{\text{-2}}}\text{ L bar }{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}\text{)(300K)}}$
Or $\text{n=}\dfrac{\text{(1}{{\text{0}}^{\text{-2}}}\text{ bar )(10 L)}}{\text{(0}\text{.083 L bar }{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}\text{)(300K)}}$
We know that an Avogadro's number is the total number of particles present per mole of the substance. The Avogadro number is equal to$\text{6}\text{.023 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{23}}}\text{mo}{{\text{l}}^{\text{-1}}}$.
Avogadro's number is related to the number of particles and the number of moles by the relation.
$\text{No}\text{.of moles=}\dfrac{\text{Number of partilcles}}{\text{Avagadro }\\!\\!'\\!\\!\text{ s no(}{{\text{N}}_{\text{A}}}\text{)}}$
We have to find out the number of helium $\text{He}$particles present in the cylinder.
$\text{Number of partilcles of He=No}\text{.of moles of He }\times \text{Avagadro }\\!\\!'\\!\\!\text{ s no(}{{\text{N}}_{\text{A}}}\text{)}$
$\text{Number of partilcles of He=}\dfrac{\text{(1}{{\text{0}}^{\text{-2}}}\text{ bar )(10 L)}}{\text{(0}\text{.083 L bar }{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}\text{)(300K)}}\text{ }\\!\\!\times\\!\\!\text{ Avagadro }\\!\\!'\\!\\!\text{ s no}\text{.(}{{\text{N}}_{\text{A}}}\text{)}$
Since Avogadro's number is$\text{6}\text{.023 }\\!\\!\times\\!\\!\text{ 1}{{\text{0}}^{\text{23}}}\text{mo}{{\text{l}}^{\text{-1}}}$. We get,
$\text{Number of partilcles of He=}\dfrac{\text{(1}{{\text{0}}^{\text{-2}}}\text{ bar )(10 L)}}{\text{(0}\text{.083 L bar }{{\text{K}}^{\text{-1}}}\text{mo}{{\text{l}}^{\text{-1}}}\text{)(300K)}}\text{ }\times \text{6}\text{.023}\times \text{1}{{\text{0}}^{\text{23}}}\text{mo}{{\text{l}}^{\text{-1}}}$
$\text{Number of partilcles of He=}\dfrac{\text{6}\text{.023}\times \text{1}{{\text{0}}^{\text{22}}}}{24.9}\text{ }$
Or $\text{Number of partilcles of He = 2}\text{.41}\times \text{1}{{\text{0}}^{\text{21}}}\text{ Atoms}$
Hence, the cylinder of capacity $\text{10L}$at $\text{300K}$contains the$\text{2}\text{.41}\times \text{1}{{\text{0}}^{\text{21}}}\text{ Atoms}$.

Hence, (C) is the correct option.

Note: The value of gas constant R depends on the unit of pressure. The value for the gas constant is as listed below,

Values of R	Units
$8.205\times {{10}^{-2}}$	$\text{L}\text{.atm}\text{.}{{\text{K}}^{\text{-1}}}\text{.mo}{{\text{l}}^{\text{-1}}}$
$8.3147\times {{10}^{-2}}$	$\text{L}\text{.bar}\text{.}{{\text{K}}^{\text{-1}}}\text{.mo}{{\text{l}}^{\text{-1}}}$
$8.314$	$\text{L}\text{.kPa}\text{.}{{\text{K}}^{\text{-1}}}\text{.mo}{{\text{l}}^{\text{-1}}}$
$8.314$	$\text{J}\text{.}{{\text{K}}^{\text{-1}}}\text{.mo}{{\text{l}}^{\text{-1}}}$
$62.364$	$\text{L}\text{.Torr}\text{.}{{\text{K}}^{\text{-1}}}\text{.mo}{{\text{l}}^{\text{-1}}}$
$1.9872$	$\text{cal}\text{.}{{\text{K}}^{\text{-1}}}\text{.mo}{{\text{l}}^{\text{-1}}}$

Use the appropriate value of the gas constant as per the requirement. Here we use the gas constant value for the pressure in the bar.