# UY1: Isothermal Process

In an isothermal process, the temperature of the system is unchanged. (ΔT = 0). For this process, the first law gives ΔE = 0, thus Q = -W.

This means that the net heat input into the system equals to the net work output by the system.

From the ideal gas equation, P1V1 = P2V2 = nRT (isotherm).

Isothermal work done by the gas,

\begin{aligned} W_{\text{out}} &= \int\limits_{V_{i}}^{V_{f}} P \, dV \\ &= \int\limits_{V_{i}}^{V_{f}} \frac{nRT}{V} \, dV \\ &= nRT \, \text{ln} \, \frac{V_{f}}{V_{i}} \end{aligned}

Note: Since PV = constant, the area of red rectangle = green rectangle. This geometric property will be useful to solve heat cycle problems.

Previous: Concept of heat engines and heat pumps

Back To Thermodynamics

##### Mini Physics

As the Administrator of Mini Physics, I possess a BSc. (Hons) in Physics. I am committed to ensuring the accuracy and quality of the content on this site. If you encounter any inaccuracies or have suggestions for enhancements, I encourage you to contact us. Your support and feedback are invaluable to us. If you appreciate the resources available on this site, kindly consider recommending Mini Physics to your friends. Together, we can foster a community passionate about Physics and continuous learning.

This site uses Akismet to reduce spam. Learn how your comment data is processed.