## delta S

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

Ajith Raja 2L
Posts: 24
Joined: Wed Nov 18, 2015 3:00 am

### delta S

If delta S equals delta H/ temperature, does that mean the more you decrease temperature the higher delta S is?

Matthew Gutierrez 2D
Posts: 19
Joined: Fri Sep 25, 2015 3:00 am

### Re: delta S

Yes, the change in entropy is greater at lower temperatures than higher temperatures.

Ajith Raja 2L
Posts: 24
Joined: Wed Nov 18, 2015 3:00 am

### Re: delta S

But I remember hearing that when temperature is zero, than delta S is also 0. Or is that wrong.

Saraid Perez 1K
Posts: 2
Joined: Tue Nov 17, 2015 3:00 am

### Re: delta S

That is right because the Third Law of Thermodynamics, states that materials have zero entropy at 0K.

GiselleMartinez_1B
Posts: 9
Joined: Mon Jan 26, 2015 2:17 pm

### Re: delta S

If you think about it algebraically it would make sense. Dividing by a smaller number vs dividing by a larger one will give you a larger value. So temperature which is the variable that will be in the denominator will determine the size of your value. Lower temp = greater change in entropy.