Change in molar entropy of steam

In summary: I'm not sure what you're doing here. You are calculating the molar entropy change in going from 373.15K to 393.15K, right? So why are you multiplying by 9/2? I don't see the 9/2 in the equation: ##dS = nC_v \ln(T_2/T_1)##I think the correct way to do this is to look up the values of ##C_p## and ##C_v## for water vapor and use the appropriate relationship between them. I think you will find that the value of ##C_v## is very close to the value for a monatomic gas, so the error of using ##C_v## for a monatomic
  • #1
says
594
12

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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  • #2
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.
 
  • #3
tnich said:
You could look up the value of ##C_v## for steam.
Oh, and check your units. Your equation is dimensionally inconsistent.
 
  • #4
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...
 
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  • #5
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
 
  • #6
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.
 
  • #7
I thought the same thing about that equation. I thought that would be how Cv is derived.
 
  • #8
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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Likes says
  • #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)
 
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  • #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?
 
  • Like
Likes says
  • #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
 

FAQ: Change in molar entropy of steam

1. What is the change in molar entropy of steam?

The change in molar entropy of steam refers to the difference in the amount of disorder or randomness of molecules in one mole of steam at two different states. It is measured in units of joules per mole-kelvin (J/mol•K).

2. How is the change in molar entropy of steam calculated?

The change in molar entropy of steam can be calculated using the formula ΔS = S2 - S1, where S1 and S2 are the entropies of the steam at two different states. The values for S1 and S2 can be found in thermodynamic tables or calculated using thermodynamic equations.

3. What factors affect the change in molar entropy of steam?

The change in molar entropy of steam is affected by temperature, pressure, and phase changes. Higher temperatures and pressures generally lead to an increase in entropy, while phase changes (such as vaporization or condensation) can cause a significant change in entropy.

4. Why is the change in molar entropy of steam important?

The change in molar entropy of steam is important because it is a measure of the amount of energy that can be converted to work in a thermodynamic process. It also helps to determine the efficiency of heat engines and other thermodynamic systems.

5. How does the change in molar entropy of steam relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. The change in molar entropy of steam is a direct measurement of this increase in entropy, as steam has a higher entropy compared to liquid water. This law also helps to explain the direction of heat flow and the irreversibility of certain processes.

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