# The Naked Scientists Forum

### Author Topic: Entropy--everything is downhill from here...  (Read 5228 times)

#### Mr Andrew

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• God was primitive man's attempt at Physics.
##### Entropy--everything is downhill from here...
« on: 28/11/2007 23:23:42 »
Where did the mathematical description of entropy come from?  I understand the qualitative idea of disorder and microstates and all of that but why does dS = dQ/T?  This equation is derived from the first law of thermodynamics for an ideal gas heated isothermically.  Why is this important?  The equation at the end of the derivation was dV/V = dQ/nRT = dS/nR where n and R are the number of moles in the gas and the gas law constant.  Why should dQ/T be entropy and not dQ/nRT?  I can't figure it out!  Can anybody else?

#### lightarrow

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##### Entropy--everything is downhill from here...
« Reply #1 on: 29/11/2007 19:36:21 »
Where did the mathematical description of entropy come from?  I understand the qualitative idea of disorder and microstates and all of that but why does dS = dQ/T?  This equation is derived from the first law of thermodynamics for an ideal gas heated isothermically.  Why is this important?  The equation at the end of the derivation was dV/V = dQ/nRT = dS/nR where n and R are the number of moles in the gas and the gas law constant.  Why should dQ/T be entropy and not dQ/nRT?  I can't figure it out!  Can anybody else?

The exact description of entropy is not simple. It's possible to show that dQR/T is a function of state, then it's possible to show that this is equal to the statistical definition of entropy S = klnW but all the demonstration are quite complex.
d = not-exact differential
QR =  reversibly exchanged heat

Simplistically we could say: we know that entropy increases in a dissipative process, that is when heat flows between to bodies which temperature difference is finite (and not infinitesimal), so dS have to be proportional to this heat; furthermore, since heat flows from a body with higher T to another with lower T, the increase in entropy of the colder body have to be greater than that of the hotter, so that the total variation of entropy is positive; so dS of a body have to be inversely (proportional) to the body's temperature: dS = dQ/T. I wrote proportional in brackets because with this reasoning we don't have proved the inverse proportionality, but only that one increases when the other decreases.

#### lyner

• Guest
##### Entropy--everything is downhill from here...
« Reply #2 on: 29/11/2007 20:58:58 »
Quote
Why should dQ/T be entropy and not dQ/nRT?
Entropy has to refer to all systems, not just an ideal gas. Perhaps. n and R are rather gas-specific.

#### Mr Andrew

• Sr. Member
• Posts: 206
• God was primitive man's attempt at Physics.
##### Entropy--everything is downhill from here...
« Reply #3 on: 30/11/2007 16:44:26 »
dQ/T = dS is derived for an isothermal process.  What if T changed and pressure and volume were contant?  Then dQ = dU and dU is proportional to T.  So, dQ/T is not proportional to T at all (T/T = 1).  If no work is done then entropy is independent of temperature?

In an adiabatic process where there is no heat then does dS = 0?  What if a gas expands adiabatically...then there IS entropy because there are more microstates possible but dS = dQ/T = 0.

Since this equation for entropy doesn't work for every process it must be specific to an isothermal process.  What might an equation be for other processes?

#### lightarrow

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##### Entropy--everything is downhill from here...
« Reply #4 on: 30/11/2007 19:18:43 »
dQ/T = dS is derived for an isothermal process.  What if T changed and pressure and volume were contant?  Then dQ = dU and dU is proportional to T.  So, dQ/T is not proportional to T at all (T/T = 1).  If no work is done then entropy is independent of temperature?

In an adiabatic process where there is no heat then does dS = 0?  What if a gas expands adiabatically...then there IS entropy because there are more microstates possible but dS = dQ/T = 0.

Since this equation for entropy doesn't work for every process it must be specific to an isothermal process.  What might an equation be for other processes?
dS = dQR/T is valid for every reversible process, so temperature doesn't need to be constant. Yes, an adiabatic process means dS = 0.

#### lightarrow

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##### Entropy--everything is downhill from here...
« Reply #5 on: 30/11/2007 19:26:10 »
Quote
Why should dQ/T be entropy and not dQ/nRT?
Entropy has to refer to all systems, not just an ideal gas. Perhaps. n and R are rather gas-specific.
Internal energy U, enthalpy H, entropy S ecc. are all extensive physical magnitudes (it's correct this term?) so they are proportional to the amount of matter; dividing them by n you have the corrisponding molar values. About R, if you do S/R then the physical dimensions don't correspond to the requested ones.

#### Mr Andrew

• Sr. Member
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• God was primitive man's attempt at Physics.
##### Entropy--everything is downhill from here...
« Reply #6 on: 01/12/2007 04:16:44 »
Why is dS = dQ/T not specific to isothermal processes?  Here is the derivation:

dU = dQ - dW = 0
dQ = dW = d(PV) = nRT
dQ/nRT = 1 = dS/nR
dS = dQ/T

I guess T doesn't have to be constant but I'm pretty sure that my book assumed an isothermally expanding ideal gas.

What kind of process is defined by dU = 0?

#### lightarrow

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##### Entropy--everything is downhill from here...
« Reply #7 on: 01/12/2007 08:22:05 »
Why is dS = dQ/T not specific to isothermal processes?  Here is the derivation:

dU = dQ - dW = 0
dQ = dW = d(PV) = nRT
No, d(PV) = nRdT

Quote
dQ/nRT = 1 = dS/nR
dS = dQ/T

I guess T doesn't have to be constant but I'm pretty sure that my book assumed an isothermally expanding ideal gas.
I haven't understood how your book does it, however dS = dQ/T is defined in that way, and for every reversible process (T constant or not).
Quote
What kind of process is defined by dU = 0?
Isothermal, at least for an ideal gas.

#### The Naked Scientists Forum

##### Entropy--everything is downhill from here...
« Reply #7 on: 01/12/2007 08:22:05 »