
Volume of the air that will be expelled from a vessel of $300\,c{{m}^{3}}$when it is heated from ${{27}^{{\mathrm O}}}C$to ${{37}^{{\mathrm O}}}C$at the same pressure will be
A.$310\,c{{m}^{3}}$
B.$290\,c{{m}^{3}}$
C.$10\,c{{m}^{3}}$
D.$37\,c{{m}^{3}}$
Answer
232.8k+ views
Hint: According to Charles’ law, the volume occupied by a fixed amount of gas molecules is directly proportional to the temperature if the pressure remains constant. Therefore by putting the given values in the mathematical expression of Charles’ law we can calculate the volume when temperature increases upon heating.
Formula Used: The mathematical expression of Charles’ law is given by,
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$
Here ${{V}_{1}}\,$and ${{V}_{2}}$denotes the initial and final volume occupied by air molecules.
${{T}_{1}}$and ${{T}_{2}}$denotes the initial and final temperature.
Complete step-by-step solution:The effect of temperature on the volume of gas at constant pressure is given by French physicist Jacques Charles. According to Charles’s law the volume,$V$ of a given mass of a gas varies directly with the absolute temperature,$T$of the gas when pressure remains constant. The absolute temperature is the temperature measured with the Kelvin scale.
$\therefore \,\,\,V\,\,\alpha \,\,T$
This law can also be applied to compare changing conditions for a gas.
Let the initial volume,${{V}_{1}}$and initial temperature,${{T}_{1}}$change under reaction conditions to the final volume,${{V}_{2}}$and final temperature,${{T}_{2}}$. Then the mathematical expression of Charles’ law,
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$ …….(i)
Given,${{V}_{1}}=300\,c{{m}^{3}}$, ${{T}_{1}}=(273+27)={{300}^{o}}K$and ${{T}_{2}}=(273+37)={{310}^{o}}K$
From (i), ${{V}_{2}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}\times {{V}_{1}}$
Or, ${{V}_{2}}=\dfrac{{{310}^{o}}K}{{{300}^{o}}K}\times 300\,c{{m}^{3}}=310\,c{{m}^{3}}$
Therefore the final volume of the air molecules is $310\,c{{m}^{3}}$.
Thus, option (A) is correct.
Note: An ideal gas always follows Charles’ law, Boyles’ law, and Avogadro's law under high temperature and low pressure. This is because the intermolecular attractions are minimal and the volume occupied by the gas molecules is negligibly smaller than the total volume of the container.
Formula Used: The mathematical expression of Charles’ law is given by,
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$
Here ${{V}_{1}}\,$and ${{V}_{2}}$denotes the initial and final volume occupied by air molecules.
${{T}_{1}}$and ${{T}_{2}}$denotes the initial and final temperature.
Complete step-by-step solution:The effect of temperature on the volume of gas at constant pressure is given by French physicist Jacques Charles. According to Charles’s law the volume,$V$ of a given mass of a gas varies directly with the absolute temperature,$T$of the gas when pressure remains constant. The absolute temperature is the temperature measured with the Kelvin scale.
$\therefore \,\,\,V\,\,\alpha \,\,T$
This law can also be applied to compare changing conditions for a gas.
Let the initial volume,${{V}_{1}}$and initial temperature,${{T}_{1}}$change under reaction conditions to the final volume,${{V}_{2}}$and final temperature,${{T}_{2}}$. Then the mathematical expression of Charles’ law,
$\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$ …….(i)
Given,${{V}_{1}}=300\,c{{m}^{3}}$, ${{T}_{1}}=(273+27)={{300}^{o}}K$and ${{T}_{2}}=(273+37)={{310}^{o}}K$
From (i), ${{V}_{2}}=\dfrac{{{T}_{2}}}{{{T}_{1}}}\times {{V}_{1}}$
Or, ${{V}_{2}}=\dfrac{{{310}^{o}}K}{{{300}^{o}}K}\times 300\,c{{m}^{3}}=310\,c{{m}^{3}}$
Therefore the final volume of the air molecules is $310\,c{{m}^{3}}$.
Thus, option (A) is correct.
Note: An ideal gas always follows Charles’ law, Boyles’ law, and Avogadro's law under high temperature and low pressure. This is because the intermolecular attractions are minimal and the volume occupied by the gas molecules is negligibly smaller than the total volume of the container.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

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

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

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

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

Equilibrium Class 11 Chemistry Chapter 6 CBSE Notes - 2025-26

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

NCERT Solutions For Class 11 Chemistry Chapter 7 Redox Reactions (2025-26)

