Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
0 votes

A gas is found to obey the law $P^3V =$ constant. The initial temperature and volume are $T_0$ and $V_0$. If the gas expands to $2 V_0$ what happens to the internal energy?

(A) increases (B) decreases (C) remains the same (D) insufficient data to resolve answer
Can you answer this question?

1 Answer

0 votes
$PV = nRT$
$P^3 V = k $ (a constant) $ \rightarrow P = (kV^{-1})^{\large\frac{1}{3}}$
$\Rightarrow (kV^{-1})^{\large\frac{1}{3}} V = nRT$
$\Rightarrow k^{\large\frac{1}{3}} V^{\large\frac{2}{3}} = nRT$
Differentiating, we get, $ k^{\large\frac{1}{3}} \large\frac{2}{3}$$ V^{\large\frac{1}{3}} dV = nRdT$
$\Rightarrow \large\frac{2}{3}$$P(\Delta V) = nR \Delta T$
Since $\Delta V$ is positive, $\Delta T$ is positive, and hence internal energy increases (i.e, it has a positive $\Delta$).
answered Jan 23, 2014 by balaji.thirumalai

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App