
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
224.7k+ views
Hint: 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,
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.
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.
Recently Updated Pages
JEE Main 2026 Session 1 Correction Window Started: Check Dates, Edit Link & Fees

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Thermodynamics Class 11 Chemistry Chapter 5 CBSE Notes - 2025-26

Organic Chemistry Some Basic Principles And Techniques Class 11 Chemistry Chapter 8 CBSE Notes - 2025-26

Hydrocarbons Class 11 Chemistry Chapter 9 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

