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 $P =kV^{-2}$. The initial volume is $V_0$. If the gas expands to $4 V_0$ and if the initial temperature is $T_0$ what is its final temperature?

(A) $2T_0$ (A) $-2T_0$ (A) $-4T_0$ (A) $4T_0$
Can you answer this question?

1 Answer

0 votes
$PV = nRT$ and $P = kV^{-2}$.
$\Rightarrow kV^{-2}V = kV^{-1} = nRT$
Differentiating, we get: $-kV^{-2} dV = nRdT \rightarrow -PdV = nRdT$
$\Rightarrow dT = \large\frac{-PdV}{nR} $$= \large\frac{-P(4V_0-V_0)}{nR} $$ = \large\frac{-nRT_0}{nRV_0}$$3V_0$ $ = -3T_0$
answered Jan 23, 2014 by balaji.thirumalai

Related questions

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