ThermodynamicsMedium
Question
A quantity of 1 g of water on evaporation at atmospheric pressure forms 1671 cm3 of steam. Heat of vaporization is 540 cal/g. The approximate increase in internal energy is (1L – atom = 24 cal)
Options
A.250 cal
B.500 cal
C.1000 cal
D.1500 cal
Solution
$q = 540\text{ cal}$
$w = - P\left( V_{\text{vap}} - V_{\text{liq}} \right) = - 1\text{ atm }(1671 - 1)\text{ c}\text{m}^{3} $$${= - \frac{1670}{1000}\text{L-atm = } - 1.670 \times 101.3J }{\text{=-}\frac{1.670 \times 101.3}{4.184}\text{ cal = } - 40\text{ cal} }{\therefore\Delta\text{U=q+w=500 cal}}$$
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