
Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure. Efficiency of this cycle is nearly (assume the gas to be close to ideal gas)

(A) $15.4\% $
(B) $9.1\% $
(C) $10.5\% $
(D) $12.5\% $
Answer
136.2k+ views
Hint: The efficiency of a cycle is equal to the ratio of the work done in that cycle to the heat absorbed. The work done can be easily evaluated by calculating the area of the P-V diagram given in the question.
Complete step-by-step solution:
We know that the work done in a process is equal to the area of the P-V diagram. So the work done in the given cycle is given by
$W = Ar\left( {ABCD} \right)$
$ \Rightarrow W = AB \times AD$
From the given diagram, we have $AB = {P_0}$ and $AD = {V_0}$. Substituting these above, we get
$W = {P_0}{V_0}$.....................(1)
Since the direction of the loop is clockwise, so the work done is positive.
Now, the heat is absorbed in the processes AB and BC. These two respectively are isochoric and isobaric processes. So we have
${Q_{AB}} = n{C_V}\Delta T$
We know that ${C_V} = \dfrac{{fR}}{2}$. So we have
${Q_{AB}} = \dfrac{{fnR\Delta T}}{2}$
$ \Rightarrow {Q_{AB}} = \dfrac{{fV\Delta P}}{2}$ (1) (from the ideal gas equation for constant pressure)
For the process AB, we have
$\Delta P = 2{P_0} - {P_0}$
$ \Rightarrow \Delta P = {P_0}$
Also $V = {V_0}$
Substituting these in (1) we get
${Q_{AB}} = \dfrac{{f{P_0}{V_0}}}{2}$....................(2)
Now, for the process BC, we have
${Q_{BC}} = n{C_P}\Delta T$
We know that ${C_P} = \left( {\dfrac{f}{2} + 1} \right)R$. So we have
${Q_{BC}} = \left( {\dfrac{f}{2} + 1} \right)nR\Delta T$
$ \Rightarrow {Q_{BC}} = \left( {\dfrac{f}{2} + 1} \right)P\Delta V$................(3) (from the ideal gas equation for constant pressure)
For the process BC we have
$\Delta V = 2{V_0} - {V_0}$
$ \Rightarrow \Delta V = {V_0}$
Also $P = 2{P_0}$
Substituting these in (3) we get
${Q_{BC}} = 2\left( {\dfrac{f}{2} + 1} \right){P_0}{V_0}$....................(4)
Adding (2) and (4) we get the total heat absorbed as
$Q = \dfrac{{f{P_0}{V_0}}}{2} + 2\left( {\dfrac{f}{2} + 1} \right){P_0}{V_0}$
$ \Rightarrow Q = {P_0}{V_0}\left[ {\dfrac{f}{2} + 2\left( {\dfrac{f}{2} + 1} \right)} \right]$
Now, since Helium is a monatomic gas, we have $f = 3$. Substituting this above, we get
$Q = {P_0}{V_0}\left[ {\dfrac{3}{2} + 2\left( {\dfrac{3}{2} + 1} \right)} \right]$
$ \Rightarrow Q = \dfrac{{13}}{2}{P_0}{V_0}$......................(5)
Now, the efficiency is given by
$\eta = \dfrac{W}{Q}$
Putting (1) and (5) we get
$\eta = \dfrac{{{P_0}{V_0}}}{{\dfrac{{13}}{2}{P_0}{V_0}}}$
$ \Rightarrow \eta = \dfrac{2}{{13}} \approx 0.154$
Multiplying by $100$, we get the percentage efficiency of the cycle equal to $15.4\% $.
Hence, the correct answer is option (1).
Note: The sign of the work done is very much important. It can be easily found out by using the direction of the loop of the cycle in a P-V diagram. A clockwise loop means a positive work, while an anticlockwise loop means a negative work.
Complete step-by-step solution:
We know that the work done in a process is equal to the area of the P-V diagram. So the work done in the given cycle is given by
$W = Ar\left( {ABCD} \right)$
$ \Rightarrow W = AB \times AD$
From the given diagram, we have $AB = {P_0}$ and $AD = {V_0}$. Substituting these above, we get
$W = {P_0}{V_0}$.....................(1)
Since the direction of the loop is clockwise, so the work done is positive.
Now, the heat is absorbed in the processes AB and BC. These two respectively are isochoric and isobaric processes. So we have
${Q_{AB}} = n{C_V}\Delta T$
We know that ${C_V} = \dfrac{{fR}}{2}$. So we have
${Q_{AB}} = \dfrac{{fnR\Delta T}}{2}$
$ \Rightarrow {Q_{AB}} = \dfrac{{fV\Delta P}}{2}$ (1) (from the ideal gas equation for constant pressure)
For the process AB, we have
$\Delta P = 2{P_0} - {P_0}$
$ \Rightarrow \Delta P = {P_0}$
Also $V = {V_0}$
Substituting these in (1) we get
${Q_{AB}} = \dfrac{{f{P_0}{V_0}}}{2}$....................(2)
Now, for the process BC, we have
${Q_{BC}} = n{C_P}\Delta T$
We know that ${C_P} = \left( {\dfrac{f}{2} + 1} \right)R$. So we have
${Q_{BC}} = \left( {\dfrac{f}{2} + 1} \right)nR\Delta T$
$ \Rightarrow {Q_{BC}} = \left( {\dfrac{f}{2} + 1} \right)P\Delta V$................(3) (from the ideal gas equation for constant pressure)
For the process BC we have
$\Delta V = 2{V_0} - {V_0}$
$ \Rightarrow \Delta V = {V_0}$
Also $P = 2{P_0}$
Substituting these in (3) we get
${Q_{BC}} = 2\left( {\dfrac{f}{2} + 1} \right){P_0}{V_0}$....................(4)
Adding (2) and (4) we get the total heat absorbed as
$Q = \dfrac{{f{P_0}{V_0}}}{2} + 2\left( {\dfrac{f}{2} + 1} \right){P_0}{V_0}$
$ \Rightarrow Q = {P_0}{V_0}\left[ {\dfrac{f}{2} + 2\left( {\dfrac{f}{2} + 1} \right)} \right]$
Now, since Helium is a monatomic gas, we have $f = 3$. Substituting this above, we get
$Q = {P_0}{V_0}\left[ {\dfrac{3}{2} + 2\left( {\dfrac{3}{2} + 1} \right)} \right]$
$ \Rightarrow Q = \dfrac{{13}}{2}{P_0}{V_0}$......................(5)
Now, the efficiency is given by
$\eta = \dfrac{W}{Q}$
Putting (1) and (5) we get
$\eta = \dfrac{{{P_0}{V_0}}}{{\dfrac{{13}}{2}{P_0}{V_0}}}$
$ \Rightarrow \eta = \dfrac{2}{{13}} \approx 0.154$
Multiplying by $100$, we get the percentage efficiency of the cycle equal to $15.4\% $.
Hence, the correct answer is option (1).
Note: The sign of the work done is very much important. It can be easily found out by using the direction of the loop of the cycle in a P-V diagram. A clockwise loop means a positive work, while an anticlockwise loop means a negative work.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs

JEE Main 2021 July 25 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 20 Shift 2 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

How to find Oxidation Number - Important Concepts for JEE

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

A body crosses the topmost point of a vertical circle class 11 physics JEE_Main

JEE Main 2025: Derivation of Equation of Trajectory in Physics

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

Degree of Dissociation and Its Formula With Solved Example for JEE

A body is falling from a height h After it has fallen class 11 physics JEE_Main

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

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
