In example 6 5 of ncert on pg 161, In both cases there is no change in volume - Chemistry - Thermodynamics

Question

In example 6 5 of ncert on pg 161,
In both cases there is no change in volume Then why for
conversion to gas we use formula delta U + nRt
And for conversion to ice we assume nRt =0?
Problem   6 5 If   water   vapour   is  
assumed   to   be   a   perfect   gas ,
molar   enthalp   change   for   vapourisation
  of 1 mol   of   water   at   1 bar
  and   100 ° C   is   41 kJ   mol - 1
Calculate   the   internal   energy   change ,
when ( i )   1   mol   of   water   is
  vaporised   at   1   bar   pressure
  and   100 ° C ( ii )   1   mol   of
  water   is   converted   into   ice
Solution :   ( i )   The   change   H 2 O ( l )
→ H 2 O ( g ) △ H = △ U + △ n g RT or  
△ U = △ H - △ n g RT , substitutinh   the
values ,   we   get △ U = 41 00   kJ  
mol - 1   - 1              
        × 8 3   J   mol - 1 K -
1 × 373 K = 41 00   kJ   mol - 1   - 3 096
  kJ   mol - 1 = 37 904   kJ   mol - 1 ( ii )
  The   change   H 2 O ( l ) → H 2 O ( s )
There   is   negligible   change   in  
volume , So ,   we   can   put   p △ V =
△ n g RT ≈ 0   in   this case , △ H
≅ △ U so ,   △ U = 41 00   kJ   mol
- 1

Bharti Paliwal · Accepted Answer

Dear Student

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