For an isothermal reversible expansion process,the value of $q$ can be calculated by the expression
- A$q = 2.303 \ nRT \ \log \ \frac{V_2}{V_1}$
- B$q = -2.303 \ nRT \ \log \ \frac{V_2}{V_1}$
- C$q = - 2.303 \ nR \ \log \ \frac{V_1}{V_2}$
- D$q = - P_{\text{exp}} \ nRT \ \log \ \frac{V_1}{V_2}$
Explore More
Similar Questions
$q$ and $w$ are both ....... functions and $q + w$ is a ....... function.
Easy
View SolutionIf $3 \ mol$ of an ideal gas at $300 \ K$ expand isothermally from $30 \ dm^3$ to $45 \ dm^3$ against a constant opposing pressure of $80 \ kPa$,then the amount of heat transferred is . . . . . . $J$.
DifficultJEE MAIN 2024
View SolutionBased on the first law of thermodynamics,which one of the following is correct?
Which of the following relations is correct for the internal energy change $(\Delta U)$ in a process occurring at constant volume?
Easy
View SolutionWhat is the amount of work done when $0.5 \ mol$ of methane,$CH_{4(g)}$,is subjected to combustion at $300 \ K$? (Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$)
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo