Entropy of Phase Change
Trending Questions
Q.
Calculate entropy change for vaporization of 1 mole of liquid water to steam at 100∘C if ΔHv=40.8 kJ mol−1.
ΔSv=107.38 JK−1 mol−1
ΔSv=109.38 JK−1 mol−1
ΔSv=109.48 JK−1 mol−1
ΔSv=107.38 JK−1 mol−1
Q. At 373 K, steam and water are in equilibrium and △vapH=40.98 kJ mol−1. What will be entropy change for conversion of water into steam?
H2O(l)→H2O(g)
H2O(l)→H2O(g)
- 109.8 J K−1 mol−1
- 31 J K−1 mol−1
- 21.98 J K−1 mol−1
- 326 J K−1 mol−1
Q. The entropy change involved in the conversion of 1 mole of liquid water at 373 K to vapour will be:
Given: △Hvap=2.257 kJ/g
Given: △Hvap=2.257 kJ/g
- 118.5 JK−1mol−1
- 108.9 JK−1mol−1
- 150 JK−1mol−1
- 130.6 JK−1mol−1
Q. Entropy change involved in conversion of one mole of liquid water at 373 K to vapour at the same temperature (latent heat of vaporisation of water =2.257 kJg−1).
- 30.7 JK−1mol−1
- 60.3 JK−1mol−1
- 9.8 JK−1mol−1
- 108.9 JK−1mol−1
Q.
Calculate entropy change for vaporization of 1 mole of liquid water to steam at 100∘C if ΔHv=40.8 kJ mol−1.
ΔSv=107.38 JK−1 mol−1
ΔSv=109.48 JK−1 mol−1
ΔSv=109.38 JK−1 mol−1
ΔSv=107.38 JK−1 mol−1
Q. Total work done on the gas in two stage compression is:
- 40
- 80
- 160
- none of these
Q. If △Hvapourisation of substance X(l) (molar mass = 30 g/mol) is 300 J/g at its boiling point 300 K, molar entropy change for reversible condensation process is:
- 30J/molK−1
- −60 J/molK−1
- −30 J/molK−1
- None of these
