Request a call back

A swimmer coming out from a pool with a film of water weighing about 18 g .how much heat must be supplied to evapourite this water?also calculaye internal energy of vapiurisation at 100℃ given Enthalpy of vapiurisation at 373k is 40.66kj/mole
Asked by ashutosharnold1998 | 11 Aug, 2019, 12:37: AM
Given:

Mass of water = 18 gm

ΔvapH2O = 40.66 kJ/mol

R = 8.314×10−3 kJ/mol

T = 373 K

Evaporation of water is given as,

ΔvapU = ΔH − PΔV

=ΔH −ΔngRT

Δng = 1−0
= 1 mol

Therefore

ΔvapU = 40.66 − (1)(8.314×10−3)(373)

= 40.66 − 3.10

ΔvapU = 37.55 kJ/mol

Internal energy of vaporisation is 37.55 kJ/mol
Answered by Varsha | 12 Aug, 2019, 11:19: AM

## Concept Videos

JEE main - Chemistry
Asked by mp0985797 | 01 Feb, 2022, 08:38: PM
JEE main - Chemistry
JEE main - Chemistry
Asked by ashutosharnold1998 | 03 Nov, 2019, 08:22: PM
JEE main - Chemistry
Asked by ashutosharnold1998 | 31 Oct, 2019, 07:16: PM
JEE main - Chemistry
Asked by ashutosharnold1998 | 11 Aug, 2019, 12:37: AM
JEE main - Chemistry
Asked by ashutosharnold1998 | 10 Aug, 2019, 12:10: AM