# A problem from thermodynamics -- Freezing of water at 273 K and 1 atm

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1. Apr 27, 2016

### vijayramakrishnan

1. The problem statement, all variables and given/known data
Freezing of water at 273 K and 1 atm

which of the following is true for the above thermodynamics process
p) q=0
q)w=0
r)ΔSsys<0
s)ΔU=0
t)ΔG=0

2. Relevant equations
none

3. The attempt at a solution

i got r, s ,t

since the reaction happens at constant temperature,internal energy is constant
since the reaction is open so it is isobaric in nature and since the volume changes due to phase change,so work done is not zero,so heat must be exchanged(first law of thermodynamics).
randomness decreases so ΔSsys<0

but answer given is r t q

2. Apr 28, 2016

### Andrew Mason

$\Delta U$ includes potential energy. In order to form ice, the water molecules lose potential energy and rotational kinetic energy. Average translational kinetic energy does not change (i.e. temperature) but average potential energy per molecule decreases and there is loss of rotational kinetic energy. The process is exothermic (heat flows out of the water to form ice).

Since $\Delta Q < 0$, $\Delta S = \int dQ/T = \Delta Q/T < 0$.

Since P and T are constant, $\Delta G = \Delta Q - T\Delta S = T\Delta S - T\Delta S = 0$

I am not sure about w = 0, however. Ice takes up more volume than liquid water (which is why ice floats). So there is a small amount of work done on the surroundings.

[later edits are in italics]

AM

Last edited: Apr 28, 2016
3. Apr 28, 2016

### vijayramakrishnan

thank you very much sir,understood it.

4. Apr 28, 2016

### Andrew Mason

With respect to the internal energy, I should have said that the average translational kinetic energy does not change. The specific heat of water is greater than the specific heat of ice. This is because H2O molecules lose rotational kinetic energy when they freeze to form ice because they lose the rotational degree of freedom.

AM