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WBJEEMatch the thermodynamic processes given under Column I with the expressions given under...
Match the thermodynamic processes given under Column I with the expressions given under Column II.
| Column I | Column II | ||
| A. | Freezing of water at 273 K and 1 atm | P. | q=0 |
| B. | Expansion of 1 mol of an ideal gas into a vacuum under isolated conditions | Q. | w=0 |
| C. | Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container | R. | ΔSsys<0 |
| D. | Reversible heating of H2(g) at 1 atm from 300 K to 600 K, followed by reversible cooling to 300 K at 1 atm | S. | ΔU=0 |
| T. | ΔG=0 |
Solution:
A - H2O(l)→H2O(s) at 273 K. & 1 atm
∆H=-ve=q
∆Ssys<0, ∆G=0
w≠0 (as water expands on freezing), ∆U≠0
B - Free expansion of ideal gas.
q = 0
w = 0
∆U=0
∆Ssys>0
∆G<0
C - Mixing of equal volume of ideal gases at constant pressure & temp in an isolated container
q = 0, w = 0, ∆U=0 , ∆Ssys>0 , ∆G<0
D - H2(g)300 KReversible⟶Heating, 1 atm600 K Reversible⟶Cooling, 1 atm300 K
q = 0, w = 0, ∆U=0 , ∆G=0 , ∆Ssys=0
∆H=-ve=q
∆Ssys<0, ∆G=0
w≠0 (as water expands on freezing), ∆U≠0
B - Free expansion of ideal gas.
q = 0
w = 0
∆U=0
∆Ssys>0
∆G<0
C - Mixing of equal volume of ideal gases at constant pressure & temp in an isolated container
q = 0, w = 0, ∆U=0 , ∆Ssys>0 , ∆G<0
D - H2(g)300 KReversible⟶Heating, 1 atm600 K Reversible⟶Cooling, 1 atm300 K
q = 0, w = 0, ∆U=0 , ∆G=0 , ∆Ssys=0
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