Change in molar entropy of steam

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Homework Help Overview

The discussion revolves around calculating the change in molar entropy of steam when heated from 100 to 120 °C at constant volume, assuming ideal gas behavior. Participants are exploring the implications of the heat capacity at constant volume (Cv) and its value for steam.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to apply the formula for change in entropy, questioning the value of Cv for steam and its implications on the calculations. There are discussions about the dimensional consistency of the equations used and the correct interpretation of heat capacities.

Discussion Status

Multiple interpretations of Cv are being explored, with some participants suggesting values based on theoretical derivations while others reference empirical data. There is an ongoing examination of the assumptions regarding the nature of steam as a polyatomic gas and the relevance of vibrational modes in determining Cv.

Contextual Notes

Participants note the lack of explicit information regarding Cv in the original problem statement, leading to varied approaches in deriving its value. There is also mention of the need to compare calculated values of specific heat capacity with authoritative sources.

says
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Homework Statement


Calculate the change in molar entropy of steam heated from 100 to 120 °C at constant volume in units J/K/mol (assume ideal gas behaviour).

Homework Equations


dS = n Cv ln(T1/T0)
T: absolute temperature

The Attempt at a Solution


100 C = 373.15 K
120 C = 393.15 K

dS = nCvln (393.15 / 373.15) = nCv0.052 J/K/mol

I don't really understand why I wasn't told what the Cv was though...
 
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says said:

Homework Statement


Calculate the change in molar entropy of steam heated from 100 to 120 °C at constant volume in units J/K/mol (assume ideal gas behaviour).

Homework Equations


dS = n Cv ln(T1/T0)
T: absolute temperature

The Attempt at a Solution


100 C = 373.15 K
120 C = 393.15 K

dS = nCvln (393.15 / 373.15) = nCv0.052 J/K/mol

I don't really understand why I wasn't told what the Cv was though...
You could look up the value of ##C_v## for steam.
 
tnich said:
You could look up the value of ##C_v## for steam.
Oh, and check your units. Your equation is dimensionally inconsistent.
 
I've tried two ways to calculate this, but have two different answers..

dS = nCvln(T1/T0)

Cv for steam = 27.5 J / mol*K
n = 1 mole
T0: 373.15 K
T1: 393.15 K

dS = (1 mol)(27.5 J / mol*K)(ln(393.15/373.15))
dS = 1.4358 J/K

This link I found says the units for molar entropy are J/mol*K though so I'm not sure if my answer is correct. https://en.wikipedia.org/wiki/Standard_molar_entropy

Cv = (∂U/∂T)V = ∂/∂T(3NkbT/2)=(3Nkb)/2

dS = (3Nkb)/2)(ln(393.15/373.15))
dS = 1.08*10-24 J/K

units are correct in both - not really sure what I'm missing here...
 
Last edited:
Ok, I think I know what I did wrong. I was getting my n and N mixed up.

dS = Cv ln(T1/T0)

Cv = (∂U/∂T)V = ∂/∂T (3NkbT/2) = 3Nkb/2 = 3nR/2

Cv,m = molar heat capacity, units J/mol*K

Cv,m = Cv/n = 3nR/2n = 3R/2

Cv,m = 3 (8.3144621) / 2 = 12.47169315 J/mol*K

dS = 12.47169315*ln(393.15/373.15) = 0.651 J/mol*K
 
says said:
Cv = (∂U/∂T)V = ∂/∂T (3NkbT/2) = 3Nkb/2 = 3nR/2
I can't help thinking that you have made a mistake in applying this equation. This would give the same molar specific heat for any substance in any phase.

I believe you had it right before when you said.
says said:
dS = nCvln(T1/T0)

Cv for steam = 27.5 J / mol*K
n = 1 mole
T0: 373.15 K
T1: 393.15 K

dS = (1 mol)(27.5 J / mol*K)(ln(393.15/373.15))
dS = 1.4358 J/K

My comment about the units was meant to point out that molar heat capacity should be
dS = Cvln (393.15 / 373.15) without the n, if Cv is molar heat capacity.
The equation as you stated it, dS = nCvln (393.15 / 373.15) would have units of J/K. You have solved that problem by setting n = 1.
 
I thought the same thing about that equation. I thought that would be how Cv is derived.
 
ΔS = Cvln(393.15/373.15) = change in molar entropy. Units are J/mol*K

Now I just need to find/derive an acceptable Cv
 
  • #10
Cv = (∂U/∂T)V = ∂/∂T (3NkbT/2) = 3Nkb/2 = 3R/2

I think that is correct
 
  • #11
says said:
Cv = (∂U/∂T)V = ∂/∂T (3NkbT/2) = 3Nkb/2 = 3R/2

I think that is correct
This formula is for a monatomic gas. Is steam monatomic?
 
  • #12
says said:
I've tried two ways to calculate this, but have two different answers..

dS = nCvln(T1/T0)

Cv for steam = 27.5 J / mol*K
n = 1 mole
T0: 373.15 K
T1: 393.15 K

dS = (1 mol)(27.5 J / mol*K)(ln(393.15/373.15))
dS = 1.4358 J/K

This link I found says the units for molar entropy are J/mol*K though so I'm not sure if my answer is correct. https://en.wikipedia.org/wiki/Standard_molar_entropy

Cv = (∂U/∂T)V = ∂/∂T(3NkbT/2)=(3Nkb)/2

dS = (3Nkb)/2)(ln(393.15/373.15))
dS = 1.08*10-24 J/K

units are correct in both - not really sure what I'm missing here...
This would be correct only if steam (water vapor) were a monoatomic gas, which (of course) it isn't.
 
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  • #13
Thanks, missed that. From what I've read steam is a polyatomic molecule, so instead of (3/2)R it should be (5/2)R
 
  • #14
Sorry, that should be 6/2 as steam has 6 degrees of freedom (ignoring vibrational motion)
 
Last edited:
  • #15
says said:
Sorry, that should be 6/2 as steam has 6 degrees of freedom (ignoring vibrational motion)
Why would you ignore vibrational motion?
 
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  • #16
Temperature isn't very high, so I thought it would be negligible. If I include it then Cv = 9R/2
 
  • #17
ΔS=(9/2)(R)*ln(393.15/373.15) = 1.953 J/mol*k
 
  • #18
says said:
Temperature isn't very high, so I thought it would be negligible. If I include it then Cv = 9R/2
Before you jump to a conclusion, you might compare your value of specific heat capacity to measured values from authoritative sources. I think you are right to consider the temperature, and to wonder how much a specific mode might be excited at that temperature.
says said:
ΔS=(9/2)(R)*ln(393.15/373.15) = 1.953 J/mol*k
 

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