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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Calculate Q for isothermal reversible expansion of one mole of ideal gas from initial pressure of 1 bar to 0.1 bar at a constant temperature of 273 K.

$\begin {array} {1 1} (A)\;-522.716 \: J & \quad (B)\;522.716\: J \\ (C)\; -5227.16\: J & \quad (D)\;5227.16 \: J \end {array}$

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For isothermal reversible expansion of ideal gas
$W = \large\frac{-2.303nRTlogP_1}{P_2}$$ = -2.303 \times1 \times 8.314 \times 273log \bigg(\large\frac{1}{0.1} \bigg)$
$= -5227.16\: J$
$Q = -W = -(-5227.16) = 5227.16\: J$
Ans : (D)
answered Mar 17, 2014 by thanvigandhi_1
 

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