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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) $4T_0$

1 Answer

$PV = nRT$ 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

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