# 2nd law of thermodynamics

1. Dec 17, 2006

### erty

According to the 2nd law of thermodynamics, a spontaneous process will occur (or rather: is very likely to occur) if $$\Delta S > 0$$.

A chemical process occurs if $$\Delta G < 0$$, where $$G = H - TS$$.

Example:
H = -100 kJ
T = 1 K
S = -10 kJ/K
so $$\Delta G = - 190 kJ$$. In this example, $$\Delta G < 0$$ but $$\Delta S < 0$$.
Doesn't this contradict the 2nd law?

2. Dec 17, 2006

### dextercioby

I don't see an S_{1} and an S_{2} so how did you compute \Delta S ?

Daniel.

3. Dec 17, 2006

### erty

Okay, then
$$\Delta H = -100 \mbox{kJ}$$
$$T = 1 \mbox{K}$$ and
$$\Delta S = -10 \mbox{kJ/K}$$
because $$\Delta G = \Delta H - T\Delta S$$ where T is constant.

4. Dec 17, 2006

### vanesch

Staff Emeritus
If H changes, then that means that the system is not left to itself but is externally acted upon. So we are not dealing with a spontaneous process.

5. Dec 17, 2006

### erty

If $$\Delta H < 0$$, then enthalpy is released from the system to the surroundings. It would make sense, if I had to add enthalpy ($$\Delta H > 0, \qquad H = U + pV$$) and the entropy decreased, but that's not true according to the Gibbs energy.

Last edited: Dec 17, 2006
6. Dec 17, 2006

### vivesdn

If the system is isolated, a spontaneous process will increase its entropy. The Universe is an isolated system in itself, so entropy is always increasing.
But if there is a variation in entalpy (a decrease) at constant temperature T, for sure your system is not isolated. The environment is absorbing such energy, increasing its entropy in a greater extent than the entropy lost by the system (the variation, in absolute value, of the entalpy is grater than that of the entropy, as variation og G is negative). So the universe, the system and the environment, experiments an increase in entropy. This is the result of the second law. No contradiction at all.

7. Dec 17, 2006

### vanesch

Staff Emeritus
No of course not. Take water. You extract enthalpy (you cool it). It freezes, and its entropy decreases...

8. Dec 18, 2006

### erty

I get it, thanks! $$\Delta S_{total} > 0$$, even though $$\Delta S_{sys} < 0$$, because $$\Delta H_{sys} < 0$$ according to the Gibbs energy and the conditions for a spontaneous process.

Is it possible to calculate the $$\Delta S_{surroundings}$$, if I know the $$\Delta H_{sys}$$? I mean, is there any proportionality between those two variables?

9. Dec 18, 2006

### erty

Yes, but the $$\Delta S_{total} > 0$$, because the surroundings get warmer.
(Or did I get this wrong?)

Last edited: Dec 18, 2006
10. Dec 18, 2006

### vivesdn

If my memory is still readable, I would say that $$\Delta S_{surroundings}$$ is equal to $$\Delta H_{sys}$$ if volume is constant (more generally, if there is no work performed of PV type). Then $$\Delta H_{sys}$$ is the thermal energy transferred.
If there is work performed, then enthalpy is not equal to Q, the heat.