Calculate the work done when `1.0` mol of water at `373K` vaporises against an atmosheric pressure of `1.0atm`. Assume ideal gas behaviour.

Options

A. `- 6200 J`
B. `-306 J`
C. `-3100 J`
D. `-1550 J`

Answer

C

Explanation

Volume of 1 mole of `H_2O` at 373 K and 1 atm
`V_1 = 18 cm ^(3) =0.018 L`
volume of 1 mole of `H_2O` vapour at 373 K and 1 atm `(PV_2 = RT)`
`V_2 = (RT)/(P)= ( 0.0821 xx 373 )/1 = 30.62 Lit `
` W= p (V_2 -V_1)=-1 (30 .60-0.018)=- 30.6 atm`
`=- 30.6 xx 101.3 J =- 455 j`
`Delta U = q+ W= 250 -455 =- 205 J`