
One mole of air $\left( {{C_v} = \dfrac{{5R}}{2}} \right)$ is confined at atmospheric pressure in a cylinder with a piston at\[{0^o}C\] . The initial volume occupied by the gas is\[\;V\] . After the equivalent of 13200J of heat is transferred to it, the volume of gas V is nearly \[\left( {1{\text{ }}atm = {{10}^5}N{m^{ - 2}}} \right)\] :
(A) 37 L
(B) 22 L
(C) 60 L
(D) 30 L
Answer
145.5k+ views
Hint: Use the ideal gas equation $PV = nRT$ to find the initial volume. Apply the first law of thermodynamics of heat transfer at constant pressure $Q = n{C_P}({T_2} - {T_1})$ and calculate the final temperature. Now, substitute the known data in Charles’ law expression $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$ to find the final volume.
Complete step-by-step solution
The equation which relates the pressure , volume and temperature of the given state of an ideal gas is known as the ideal gas equation or equation of state.
$PV = nRT$
Given that
\[
{C_v} = 2.5R \\
{T_1} = {\text{ }}{0^o}C{\text{ }} = 273.15K \\
P = 1{\text{ }}atm \\
\]
Substitute in the ideal gas equation.
$
PV = nR{T_1} \\
1 \times V = 1 \times 0.082 \times 273.15 \\
V = 22.4L \\
$
We know that,
$
{C_P} = R + {C_V} \\
{C_P} = R + 2.5R \\
{C_P} = 3.5R \\
$
The heat transferred Q at constant pressure is 13200J.
From first law of thermodynamics,
$
Q = n{C_P}({T_2} - {T_1}) \\
13200 = 1 \times 3.5 \times 8.314 \times ({T_2} - 273.15) \\
{T_2} = 726.77K \\
$
\[{V_1}, {\text{ }}{V_2}\] and \[{T_1},{\text{ }}{T_2}\] are the initial and final volume and temperature respectively.
According to Charles law, for the pressure remaining constant, the volume of the given mass of a gas is directly proportional to its absolute temperature.
$
V \propto T \\
\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}} \\
{V_2} = \dfrac{{726.77 \times 22.4}}{{273.15}} \\
{V_2} = 60L \\
$
Hence, the final volume is 60 L and the correct option is C.
Note: Boyle’s law states that for a given mass of an ideal gas at constant temperature , the volume is inversely proportional to the pressure.
${P_1}{V_1} = {P_2}{V_2}$
Gay-Lussac’s law states that the volume remaining constant, the pressure of a given mass of a gas is directly proportional to its absolute temperature.
Complete step-by-step solution
The equation which relates the pressure , volume and temperature of the given state of an ideal gas is known as the ideal gas equation or equation of state.
$PV = nRT$
Given that
\[
{C_v} = 2.5R \\
{T_1} = {\text{ }}{0^o}C{\text{ }} = 273.15K \\
P = 1{\text{ }}atm \\
\]
Substitute in the ideal gas equation.
$
PV = nR{T_1} \\
1 \times V = 1 \times 0.082 \times 273.15 \\
V = 22.4L \\
$
We know that,
$
{C_P} = R + {C_V} \\
{C_P} = R + 2.5R \\
{C_P} = 3.5R \\
$
The heat transferred Q at constant pressure is 13200J.
From first law of thermodynamics,
$
Q = n{C_P}({T_2} - {T_1}) \\
13200 = 1 \times 3.5 \times 8.314 \times ({T_2} - 273.15) \\
{T_2} = 726.77K \\
$
\[{V_1}, {\text{ }}{V_2}\] and \[{T_1},{\text{ }}{T_2}\] are the initial and final volume and temperature respectively.
According to Charles law, for the pressure remaining constant, the volume of the given mass of a gas is directly proportional to its absolute temperature.
$
V \propto T \\
\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}} \\
{V_2} = \dfrac{{726.77 \times 22.4}}{{273.15}} \\
{V_2} = 60L \\
$
Hence, the final volume is 60 L and the correct option is C.
Note: Boyle’s law states that for a given mass of an ideal gas at constant temperature , the volume is inversely proportional to the pressure.
${P_1}{V_1} = {P_2}{V_2}$
Gay-Lussac’s law states that the volume remaining constant, the pressure of a given mass of a gas is directly proportional to its absolute temperature.
Recently Updated Pages
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

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

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

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

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

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

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

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