Question

A swimmer coming out from the pool is covered with a thin film of water weighing about 108g. How much hat mat be soled evaporate this water? (Given: enthalpy change of water &H of water = 50 la mort and molar mass of water = 18 g marily vap A. O 150 B. 300 kJ C. 50 kJ D. 250 kJ​

Answer 1

Answer:

Step-by-step explanation:

10g

H  

2

​

O

​

(l)  

vaporisation

​

 

10g

H  

2

​

O

​

(g)     Δ  

vap

​

H  

⊝

=40.66kJmol  

−1

 

1mol                          1mol

As ideal gas, Δ  

vap

​

U  

⊝

=Δ  

vap

​

H  

⊝

−pΔV

                                     =Δ  

vap

​

H  

⊝

−Δn  

g

​

RT

Here Δn  

g

​

=1−0=1mol

Δ  

vap

​

U  

⊝

=40.66kJmol  

1

−(1mol)(8.314×10  

−3

kJmol  

−1

k  

1

)×(373k)

=40.66kJmol  

−1

−3.10

=37.56kJmol  

−1