
Calculate molar volume of a gas at STP.
Answer
157.2k+ views
Hint: Ideal Gas Equation is the equation defining the states of the hypothetical gases expressed mathematically by the combinations of empirical and physical constants. It is a relation between pressure, volume, number of moles, and temperature.
Complete step by step answer:
At standard Temperature and Pressure (STP) the molar volume \[\left( {{{\rm{V}}_{\rm{m}}}} \right)\] is the volume occupied by one mole of a chemical element or a chemical compound.
We know that at STP, \[{\rm{P = 1 atm}}\], \[{\rm{T = 273 K}}\], \[{\rm{n = 1 moles}}\] and\[{\rm{R = 0}}{\rm{.0821 L - atmmo}}{{\rm{l}}^{ - 1}}{\rm{ }}{{\rm{K}}^{ - 1}}\].
Ideal gas equation is given as –
\[{\rm{PV = nRT}}\]
\[{\rm{V = }}\dfrac{{{\rm{nRT}}}}{{\rm{P}}}\]
\[{\rm{ = }}\dfrac{{0.0821{\rm{ }} \times {\rm{ 273}}}}{1}\]
\[{\rm{V = 22}}{\rm{.4 L}}\]
Therefore, we have V = 22.4 L
Hence, the molar volume of a gas at STP is 22.4 litres.
Additional information:
Ideal gas equation was made by combining various gas laws such as Charles’s law, Boyle’s law, Gay-Lussac’s law and Avogadro’s law.The ideal gas law is the equation of state of a hypothetical ideal gas. It is an approximation of behaviour of gases under given conditions.
An ideal gas is a hypothetical gas consisting of randomly moving particles that undergo fully elastic collisions. Even though there is no such thing as an ideal gas, most of the gases tend to reach these properties when their density decreases. This happens because the intermolecular distances between gas molecules are so large that they do not interact with each other as such.
At extremely low densities, all real gases show ideal behaviour and tend to obey one universal law which we call as ideal gas law or ideal gas equation.
Note: Hence, in all the standard conditions of temperature, pressure and number of moles, the molar volume is 22.4 Litre which is also considered as a standard.
Complete step by step answer:
At standard Temperature and Pressure (STP) the molar volume \[\left( {{{\rm{V}}_{\rm{m}}}} \right)\] is the volume occupied by one mole of a chemical element or a chemical compound.
We know that at STP, \[{\rm{P = 1 atm}}\], \[{\rm{T = 273 K}}\], \[{\rm{n = 1 moles}}\] and\[{\rm{R = 0}}{\rm{.0821 L - atmmo}}{{\rm{l}}^{ - 1}}{\rm{ }}{{\rm{K}}^{ - 1}}\].
Ideal gas equation is given as –
\[{\rm{PV = nRT}}\]
\[{\rm{V = }}\dfrac{{{\rm{nRT}}}}{{\rm{P}}}\]
\[{\rm{ = }}\dfrac{{0.0821{\rm{ }} \times {\rm{ 273}}}}{1}\]
\[{\rm{V = 22}}{\rm{.4 L}}\]
Therefore, we have V = 22.4 L
Hence, the molar volume of a gas at STP is 22.4 litres.
Additional information:
Ideal gas equation was made by combining various gas laws such as Charles’s law, Boyle’s law, Gay-Lussac’s law and Avogadro’s law.The ideal gas law is the equation of state of a hypothetical ideal gas. It is an approximation of behaviour of gases under given conditions.
An ideal gas is a hypothetical gas consisting of randomly moving particles that undergo fully elastic collisions. Even though there is no such thing as an ideal gas, most of the gases tend to reach these properties when their density decreases. This happens because the intermolecular distances between gas molecules are so large that they do not interact with each other as such.
At extremely low densities, all real gases show ideal behaviour and tend to obey one universal law which we call as ideal gas law or ideal gas equation.
Note: Hence, in all the standard conditions of temperature, pressure and number of moles, the molar volume is 22.4 Litre which is also considered as a standard.
Recently Updated Pages
JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

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

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Displacement-Time Graph and Velocity-Time Graph for JEE

Geostationary Satellites and Geosynchronous Satellites - JEE Important Topic

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

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

JEE Advanced 2025 Notes

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
