Question

What is value of $△U$ (heat change at constant volume) for reversible isothermal evaporation of 90 g water at ${100}^{o}C.$ Assuming water vapour behaves as an ideal gas and $△H_{vap}{)}_{water}=540cal{g}^{-1}$

• A. $9\times {10}^{3}kcal$
• B. $6\times {10}^{3}kcal$
• C. $4.49\times {10}^{3}cal$
• D. none of these

Answer 1

The correct option is C $4.49\times {10}^{3}cal$

Heat of system $(△H{)}_v=m\times L_v$
$=90\times 540=48600cal$
$△H=△U+P(V_V-V_L)=△U+nRT$
or, $48600=△U+\frac{90}{18}\times 2\times 373$
or, $△U=48600-3730=44870cal$