The vapour pressure of water is 0.04atm at 27°C. The free energy change for the following process is H2Og0.04atm270C→H2Ol0.04atm270C
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a
0
b
R×300×ln125
c
R×300×ln25
d
300R
answer is A.
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Detailed Solution
Given that the vapor pressure of water at 27°C is 0.04 atm and the process is also occurring at 0.04 atm. Therefore, the system is at equilibrium.At equilibrium ΔG=0