## Calculating Boyle temperature with 2nd Virial coefficient

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### Calculating Boyle temperature with 2nd Virial coefficient

I'm making some thermodynamics assignments and I can't for the life of me figure out this one, it's extremely frustrating and I'd be really glad if someone could point out what I'm doing wrong.

Q: Of a gas the virial coefficient is defined as B2 = b - a /(RT) with a = 1.669 bar L2/mol2 and b = 0.01 L/mol. Calculate the Boyle temperature (in Kelvin).

So I thought, well that's easy you just rewrite the formula and equate it to zero: T=a/(RB) entering the values like they're given above in the formula gives an answer so wrong it's not even funny, then I thought: Obviously bar should be converted to Pascal, which gives a = 166900 Pa L2/mol2 , but that's even more wrong. Somehow.

I'm at a loss here, am I missing something? I checked all the units and as far can tell they're all correct, everything is in SI units.

The answer should be 2.007 K, according to the assignment.

Thanks!!
pietertje

Posts: 90
Joined: Sat Nov 08, 2008 12:43 pm UTC

### Re: Calculating Boyle temperature with 2nd Virial coefficien

Liter is not an SI unit. It should be in cubic metres, m^3. Everything else looks fine to me (except that you wrote B instead of b in the formula you derived).

jaap

Posts: 1771
Joined: Fri Jul 06, 2007 7:06 am UTC

### Re: Calculating Boyle temperature with 2nd Virial coefficien

jaap wrote:Liter is not an SI unit. It should be in cubic metres, m^3. Everything else looks fine to me (except that you wrote B instead of b in the formula you derived).

Well, if everything is in Liter, shouldn't it then cancel itself out?
pietertje

Posts: 90
Joined: Sat Nov 08, 2008 12:43 pm UTC

### Re: Calculating Boyle temperature with 2nd Virial coefficien

pietertje wrote:
jaap wrote:Liter is not an SI unit. It should be in cubic metres, m^3. Everything else looks fine to me (except that you wrote B instead of b in the formula you derived).

Well, if everything is in Liter, shouldn't it then cancel itself out?

No it doesn't all cancel.

R is usually in J/(K mol)
a is in Pa L^2/mol^2
b is in L/mol

This means T=a/(Rb) gives its answer in the units Pa L K / J.

Pascal is actually N/m^2, and a Joule is Nm, which leaves us with a T in the units K L / m^3,
so the answer you got still had one conversion factor from m^3 to Liter in it which did not cancel.

jaap

Posts: 1771
Joined: Fri Jul 06, 2007 7:06 am UTC

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