You could look up the value of ##C_v## for steam.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...
Oh, and check your units. Your equation is dimensionally inconsistent.tnich said:You could look up the value of ##C_v## for steam.
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.says said:Cv = (∂U/∂T)V = ∂/∂T (3NkbT/2) = 3Nkb/2 = 3nR/2
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
This formula is for a monatomic gas. Is steam monatomic?says said:Cv = (∂U/∂T)V = ∂/∂T (3NkbT/2) = 3Nkb/2 = 3R/2
I think that is correct
This would be correct only if steam (water vapor) were a monoatomic gas, which (of course) it isn't.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...
Why would you ignore vibrational motion?says said:Sorry, that should be 6/2 as steam has 6 degrees of freedom (ignoring vibrational motion)
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:Temperature isn't very high, so I thought it would be negligible. If I include it then Cv = 9R/2
says said:ΔS=(9/2)(R)*ln(393.15/373.15) = 1.953 J/mol*k
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).
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.
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.
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.
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.