I am trying to do the following problem:
Freezing of water at $273~\mathrm K$ and $1~\mathrm{atm}$
Which of the following is true for the above thermodynamics process?
p) $q=0$
q) $w=0$
r) $\Delta S_\mathrm{sys}<0$
s) $\Delta U=0$
t) $\Delta G=0$
My answer was r, s, and t:
- Since the reaction happens at constant temperature, internal energy is constant (yes for s)
- Since the reaction is open, it is isobaric in nature; and since the volume changes due to the phase change the work done is not zero (no for q)
- Heat must be exchanged (first law of thermodynamics) (no for p)
- Randomness decreases so $\Delta S_\mathrm{sys} < 0$ (yes for r)
But, the answer given is q, r, and t. I don't understand why q is correct but s is not.