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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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The value of $ \Delta E$ for reversible isothermal evaporation of $90g$ of water at $100^{\circ}C$. Assume that water vapor behaves as ideal gas and $ \Delta H_{vap}$ of water is $540\: cal/g(R = 2\: cal/molK)$

$\begin {array} {1 1} (A)\;44870 \: cal & \quad (B)\;-44870\: cal \\ (C)\;52330 \: cal & \quad (D)\;-52330\: cal \end {array}$

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The total heat change, $ \Delta H = 90 \times 540 = 48600 \: cal$
Now, $ \Delta H = \Delta E + P\Delta V$
$\Delta V = (V_{vapor} – V_{liquid}) = V_{vapor} (V_{liquid} is negligible)$
$\Delta H = \Delta E + PV_{vapor} = \Delta E + nRT$
$\Rightarrow \Delta E = \Delta H – nRT$
$ = 48600 – 90/18 \times 2 \times 373 = 48600 – 3730 = 44870 \: cal$
Ans : (A)
answered Mar 20, 2014 by thanvigandhi_1
 

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