# Calculating entropy

## Homework Statement

A gas sample containing 3.00 moles of Helium gas undergoes a state change from 30 degrees celsius and 25.0 L to 45 degrees celsius and 15.0 L. What is the entropy change for the gas (ideal gas)? For He, Cp = 20.8 J/k*mol

## Homework Equations

ΔS = Cv*ln(Tf/Ti) + nR*ln(Vf/Vi) = Cp*ln(Tf/Ti - nRln(Pf/Pi)

I don't know which equation to use ... I'm given Cp so do I have to convert it to Cv?

Cp = Cv + nR

## The Attempt at a Solution

I tried converting Cp to Cv:

Cv = (20.8J/K*mol)(3mol) - (3mol)(8.314J/K*mol)

Cv = 37.458 J/K (not J/Kmol anymore)

Then,

ΔS = Cv*ln(Tf/Ti) + nR*ln(Vf/Vi)

ΔS = (37.458)*ln(45/30) + (3)(8.314)*ln(15/25)

ΔS = 2.447 J/k

But the answer is -10.9 J/K.

What am I doing wrong? Thanks

collinsmark
Homework Helper
Gold Member
I haven't gone through the calculations but I've noticed there are at least two or three things going wrong with your methods.

You seem to be mixing and matching you use of relative vs. absolute approach.

• If you are working with relative measures, then $v$ is assumed to be "volume per mole" and $\Delta s$ is assumed to be "change in entropy per mole." In this approach, $\Delta s = C_v \ln \frac{T_f}{T_i} + R \ln \frac{v_f}{v_i}$. Notice that there is no variable $n$, since it's assumed that the entire equation is represented in "per mole."
• If you are working in absolute terms -- where you have a known amount of gas (e.g., 3.00 moles) then $V$ is simply volume and $\Delta S$ is simply change in entropy. In that case, $\Delta S = nC_v \ln \frac{T_f}{T_i} + nR \ln \frac{V_f}{V_i}$.

In either case, $C_p = C_v + R$.

Before plugging your temperatures in, you need to convert them to an absolute scale, such as Kelvin. Celsius isn't going to work here.