# One gram of water on evaporation at atmospheric pressure forms 1671cm3 of steam.heat of vaporisation at this pressure is 540 cal/gm. the increase in internal energy is....? a.250cal b.500cal 3.1000cal 4.1500cal

1
by sweetygannerla

2015-04-15T22:07:49+05:30

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Mass  m =  1 gm
pressure P = 1 atm
volume V = 1,671 cm³
Let the absolute temperature of water be = T ⁰K

Using the ideal gas law:    P V = n R T
1.013 * 10^5 * 1,671 * 10⁻⁶ = 1/18 mole * 8.314 J/⁰K/mole * T
T = 366.48⁰K

work done in expansion of the vapour from 1 cm³ water to 1671 cm³ vapour: at 1 atm. pressure:   P (V2 - V1) = 1.013 * 10⁵ N/m² * (1671-1) * 10⁻⁶ m³
= 169.17 Joules = 169.17/4.184  cal = 40.43 Cal

Latent heat of vapourization = 540 cal / gm = 2257 J/gm

The increase in the internal energy = latent heat -  work done in expansion
= 540 cal - 40.43 cal = 500 cal  nearly.