Work done by air when it expands from 50L to 150L at a constant pressure of 2 atmosphere?
A. $2 \times {10^4}J$
B. $2 \times {10^2}J$
C. $2 \times {10^{11}}J$
D. $2 \times {10^{15}}J$
Answer
300.9k+ views
Hint: In the question we have given that a gas is expanding from 50L to 150L at constant pressure. Here, the expansion of gas is done at constant pressure, thus the external work done can be easily calculated using the work done formula $W = P\Delta V$ (Pressure-volume work). Before starting with the calculations we must convert 50L and 150L in terms of the $m^3$ as the options provided here are in terms of Joules.
Formula used:
Workdone by a gas is given by:
$W=P \Delta V=P \left({V_2}-{V_1}\right)$ Where,
${V_1}$-The initial volume
${V_2}$-The final volume
Complete answer:
Since the gas expands at a constant pressure of $P = 2atm \approx 2 \times {10^5}N/{m^2}$
During the expansion of gas, the volume changes as:
${V_1} = 50L = 50 \times {10^{ - 3}}{m^3}$ and
${V_2} = 150L = 150 \times {10^{ - 3}}{m^3}$ (given)
$\left( {\therefore 1L = {{10}^{ - 3}}{m^3}} \right)$
Therefore, change in volume can be given as:
$\Delta V = \;\;{V_2} - {V_1} = \left( {150 - 50} \right){10^{ - 3}} = 100 \times {10^{ - 3}}{m^3}$
Now, we know that Pressure-Volume Work in thermodynamics is defined as:
$Workdone = W = P\Delta V$
Substituting the value of $P$ and $\Delta V$ in the above expression, we get
$ \Rightarrow W = 2 \times {10^5}\left( {100 \times {{10}^{ - 3}}} \right)$
$ \Rightarrow W = 2 \times {10^8} \times {10^{ - 3}}$
Approximating the values, we get
$ \Rightarrow W = 2 \times {10^4}J = 20KJ$
$\left( {\therefore 1KJ = {{10}^3}J} \right)$
Thus, the work done by air when it expands at constant pressure is about $2 \times {10^4}J$.
Hence, the correct solution is (A) $2 \times {10^4}J$.
Note: We know that the air is made up of constantly moving molecules. Therefore, when the air expands and becomes less compressed, the molecules need more space to move around. As a result, work is done by the molecules of air in expanding the space around each molecule which is calculated by using the above stated relation.
Formula used:
Workdone by a gas is given by:
$W=P \Delta V=P \left({V_2}-{V_1}\right)$ Where,
${V_1}$-The initial volume
${V_2}$-The final volume
Complete answer:
Since the gas expands at a constant pressure of $P = 2atm \approx 2 \times {10^5}N/{m^2}$
During the expansion of gas, the volume changes as:
${V_1} = 50L = 50 \times {10^{ - 3}}{m^3}$ and
${V_2} = 150L = 150 \times {10^{ - 3}}{m^3}$ (given)
$\left( {\therefore 1L = {{10}^{ - 3}}{m^3}} \right)$
Therefore, change in volume can be given as:
$\Delta V = \;\;{V_2} - {V_1} = \left( {150 - 50} \right){10^{ - 3}} = 100 \times {10^{ - 3}}{m^3}$
Now, we know that Pressure-Volume Work in thermodynamics is defined as:
$Workdone = W = P\Delta V$
Substituting the value of $P$ and $\Delta V$ in the above expression, we get
$ \Rightarrow W = 2 \times {10^5}\left( {100 \times {{10}^{ - 3}}} \right)$
$ \Rightarrow W = 2 \times {10^8} \times {10^{ - 3}}$
Approximating the values, we get
$ \Rightarrow W = 2 \times {10^4}J = 20KJ$
$\left( {\therefore 1KJ = {{10}^3}J} \right)$
Thus, the work done by air when it expands at constant pressure is about $2 \times {10^4}J$.
Hence, the correct solution is (A) $2 \times {10^4}J$.
Note: We know that the air is made up of constantly moving molecules. Therefore, when the air expands and becomes less compressed, the molecules need more space to move around. As a result, work is done by the molecules of air in expanding the space around each molecule which is calculated by using the above stated relation.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Understanding Uniform Acceleration in Physics

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2026-27 Free PDF Download (Sign-in Required)

