Q26 / 90 · JEE Main 2019 (09 Jan Shift 2) · Chemistry

The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is:

(Specific heat of water liquid and water vapour are 4.2 kJ K- and 2.0 kJ K-1kg-1 ; heat of liquid fusion and vaporization of water are 334 kJ kg-1 and 2491 kJ kg-1 , respectively). ( log 273=2.436,log⁡373=2.572,log⁡383=2.583 )

  • Correct answer
    A
    9.26 kJ kg-1K-1
  • B
    2.64 kJ kg-1K-1
  • C
    8.49 kJ kg-1K-1
  • D
    7.90 kJ kg-1K-1

Solution

Figure

(i ) Phase change H 2 O( l )→ H 2 O( s )

So Figure

(ii) Temperature change phase same (fusion)

H 2 O( l ) → Δ S 2 Δ H 2 O( l )

ΔS2= mS ln T2T1=1×4.2 ln 373273=1.31

(iii) Phase change (vapourisation)

ΔS3=ΔHvap373=2491373=6.67

(iv) Temperature change phase same

H2O(g)373 K→ΔS4H2O(g)383 K

ΔS4=2 ln383373=0.0529

ΔS  total=ΔS1+ΔS2+ΔS3+ΔS4=1.223+1.31+6.67+0.0529=9.26  kJ/kg K

Hence, the correct answer is option: 'A'.

Relevant Questions