Derive an expression for work done by the gas in an Isothermal process.
Answer
551.5k+ views
Hint: In an isothermal process, the temperature remains constant. When work is done, the volume expands thereby reducing the pressure. It is a thermodynamic process. In this process, the transfer of heat to the surroundings takes place to make the temperature constant.
Complete step by step solution:
Let us consider 1 mole of gas is enclosed in an isothermal container. Let ${P_1},$${V_1}$ and T be the initial pressure, initial volume, and temperature. As work is done, let the gas expand to ${P_2},$${V_2}$ where ${P_2}$ is the reduced pressure and ${V_2}$ is the expanded volume.
Since the process is an Isothermal Process, the temperature remains constant. We know that work done is given by,
$W = \int {dW} $
$ \Rightarrow W = \int_{{V_1}}^{{V_2}} {PdV} \_\_\_\_\_\_\_\_\left( 1 \right)$
We have the relation $PV = nRT$
$ \Rightarrow PV = RT$ $\left( {\because n = 1mole} \right)$ and R is the ideal gas constant.
$ \Rightarrow P = \dfrac{{RT}}{V}$
Substituting the value of P in equation 1, we get
$ \Rightarrow W = RT\int_{{V_1}}^{{V_2}} {\dfrac{{dV}}{V}} $
$ \Rightarrow W = RT\left[ {\ln V} \right]_{{V_1}}^{{V_2}}$
$ \Rightarrow W = RT\left[ {\ln {V_2} - \ln {V_1}} \right]$
\[ \Rightarrow W = RT\ln \dfrac{{{V_2}}}{{{V_1}}}\]
\[\therefore W = 2.303RT{\log _{10}}\dfrac{{{V_2}}}{{{V_1}}}\]
We know that for constant temperature,
\[\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{V_2}}}{{{V_1}}}\]
Thus, \[W = 2.303RT{\log _{10}}\dfrac{{{P_1}}}{{{P_2}}}\]
Thus, work done by the gas in an isothermal process is given by the expression, \[W = 2.303RT{\log _{10}}\dfrac{{{P_1}}}{{{P_2}}}\].
Note: 1. We can say that the work is positive when the force and the displacement are in the same direction and work is negative when the force and the displacement are opposite in direction. When the force and the displacement are perpendicular to each other, then the work done is zero or zero work.
2. During the isothermal process, both pressure and volume changes. Some heat engines like the Carnot Cycle are carried out by the isothermal process. Also, this process is of special interest to the ideal gases.
3. In the adiabatic process, the system does not exchange any heat with the surroundings, which is a contrast to the isothermal process.
Complete step by step solution:
Let us consider 1 mole of gas is enclosed in an isothermal container. Let ${P_1},$${V_1}$ and T be the initial pressure, initial volume, and temperature. As work is done, let the gas expand to ${P_2},$${V_2}$ where ${P_2}$ is the reduced pressure and ${V_2}$ is the expanded volume.
Since the process is an Isothermal Process, the temperature remains constant. We know that work done is given by,
$W = \int {dW} $
$ \Rightarrow W = \int_{{V_1}}^{{V_2}} {PdV} \_\_\_\_\_\_\_\_\left( 1 \right)$
We have the relation $PV = nRT$
$ \Rightarrow PV = RT$ $\left( {\because n = 1mole} \right)$ and R is the ideal gas constant.
$ \Rightarrow P = \dfrac{{RT}}{V}$
Substituting the value of P in equation 1, we get
$ \Rightarrow W = RT\int_{{V_1}}^{{V_2}} {\dfrac{{dV}}{V}} $
$ \Rightarrow W = RT\left[ {\ln V} \right]_{{V_1}}^{{V_2}}$
$ \Rightarrow W = RT\left[ {\ln {V_2} - \ln {V_1}} \right]$
\[ \Rightarrow W = RT\ln \dfrac{{{V_2}}}{{{V_1}}}\]
\[\therefore W = 2.303RT{\log _{10}}\dfrac{{{V_2}}}{{{V_1}}}\]
We know that for constant temperature,
\[\dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{{V_2}}}{{{V_1}}}\]
Thus, \[W = 2.303RT{\log _{10}}\dfrac{{{P_1}}}{{{P_2}}}\]
Thus, work done by the gas in an isothermal process is given by the expression, \[W = 2.303RT{\log _{10}}\dfrac{{{P_1}}}{{{P_2}}}\].
Note: 1. We can say that the work is positive when the force and the displacement are in the same direction and work is negative when the force and the displacement are opposite in direction. When the force and the displacement are perpendicular to each other, then the work done is zero or zero work.
2. During the isothermal process, both pressure and volume changes. Some heat engines like the Carnot Cycle are carried out by the isothermal process. Also, this process is of special interest to the ideal gases.
3. In the adiabatic process, the system does not exchange any heat with the surroundings, which is a contrast to the isothermal process.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

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

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

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

JEE Main Colleges 2026: Complete List of Participating Institutes

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

Hybridisation in Chemistry – Concept, Types & Applications

Other Pages
JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

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

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

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

