Change in molar entropy of steam

[says]

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 C_v ln(T₁/T₀)
T: absolute temperature

The Attempt at a Solution

100 C = 373.15 K
120 C = 393.15 K

dS = nC_vln (393.15 / 373.15) = nC_v0.052 J/K/mol

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

 

[tnich]

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 C_v ln(T₁/T₀)
T: absolute temperature

The Attempt at a Solution

100 C = 373.15 K
120 C = 393.15 K

dS = nC_vln (393.15 / 373.15) = nC_v0.052 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]

You could look up the value of $C_v$ for steam.

Oh, and check your units. Your equation is dimensionally inconsistent.

 

[says]

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

C_v = (∂U/∂T)_V = ∂/∂T(3Nk_bT/2)=(3Nk_b)/2

dS = (3Nk_b)/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...

 

[says]

Ok, I think I know what I did wrong. I was getting my n and N mixed up.

dS = C_v ln(T₁/T₀)

C_v = (∂U/∂T)_V = ∂/∂T (3Nk_bT/2) = 3Nk_b/2 = 3nR/2

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

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

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

dS = 12.47169315*ln(393.15/373.15) = 0.651 J/mol*K

 

[tnich]

C_v = (∂U/∂T)_V = ∂/∂T (3Nk_bT/2) = 3Nk_b/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.

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 = C_vln (393.15 / 373.15) without the n, if C_v is molar heat capacity.
The equation as you stated it, dS = nC_vln (393.15 / 373.15) would have units of J/K. You have solved that problem by setting n = 1.

 

[says]

I thought the same thing about that equation. I thought that would be how C_v is derived.

 

[says]

ΔS = C_vln(393.15/373.15) = change in molar entropy. Units are J/mol*K

Now I just need to find/derive an acceptable C_v

 

[says]

http://physics.bu.edu/~redner/211-sp06/class-thermodynamics/heatcap_volume.html

 

[says]

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

I think that is correct

 

[tnich]

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?

 

[Chestermiller]

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

C_v = (∂U/∂T)_V = ∂/∂T(3Nk_bT/2)=(3Nk_b)/2

dS = (3Nk_b)/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.

 

[says]

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

 

[says]

Sorry, that should be 6/2 as steam has 6 degrees of freedom (ignoring vibrational motion)

 

[tnich]

Sorry, that should be 6/2 as steam has 6 degrees of freedom (ignoring vibrational motion)

Why would you ignore vibrational motion?

 

[says]

Temperature isn't very high, so I thought it would be negligible. If I include it then C_v = 9R/2

 

[says]

ΔS=(9/2)(R)*ln(393.15/373.15) = 1.953 J/mol*k

 

[tnich]

Temperature isn't very high, so I thought it would be negligible. If I include it then C_v = 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.

ΔS=(9/2)(R)*ln(393.15/373.15) = 1.953 J/mol*k