Question

What is w when 2.87 kg of H₂O(l), initially at 25.0ºC, is converted into water vapour at 157 ºC against a constant external pressure of 1.00 atm? Assume that the vapour behaves ideally and that the density of liquid water is 1.00 g/mL. (Remember to include a "+" or "−" sign as appropriate.)

Answer 1

Sol.

As Mass of water = 2.87 Kg = 2870 g

Molar Mass of water = 18.01 g/mol

So , Moles of water = n = 2870 / 18.01 = 159.3559 mol  

Gas constant = R = 0.0821 L atm / K mol

External Pressure = P = 1 atm

Temperature of liquid water = T1 = 25°C = 298 K

Temperature of water vapour = T2 = 157°C = 430 K

So , Volume of liquid water = V1 = nRT1 / P

and Volume of water vapour = V2 = nRT2 / P

Therefore ,  

Work done  

= - P ( V2 - V1 )

= - P ( nRT2 / P - nRT1 / P )  

= - P ( nR / P ) ( T2 - T1 )   

= - nR ( T2 - T1 )

= - 159.3559 × 0.0821 × ( 430 - 298 )

= - 1726.9717 L atm

= - 1726.9717 × 101.325 J

=   - 174985.407 J