[orgo814]

"A sample of 70 mmol Kr(g) expands reversibly and isothermally at 373 K from 5.25 cm^3 to 6.29 cm^3, and the internal energy of the sample is known to increase by 83.5 J. Use the virial equation of state up to the second coefficient B= -28.7 cm^3 mol^-1 to calculate w, q, and delta H for this change in state."

My reasoning was this. For work = -PDeltaV, I derived that Work = -RT(1+ B/delta Vm) since the virial equation of state is PVm= RT(1 + B/Vm). Once I got work, I could calculate q since delta U (given) is equal to Q + W. For delta H, since H = U + PV, I would have delta H = delta U + RT(1 + B/delta Vm). If you could tell me if my reasoning is correct, that would be great. If it is not, please let me know what I did wrong and how to correct that.

Also, for the value of R in that equation, would I use 0.08206 or 8.314? Since I'm in cm^3 and they're both in units dm^3 or m^3.

Thanks for any help in advance!

[MrTeo]

"A sample of 70 mmol Kr(g) expands reversibly and isothermally at 373 K from 5.25 cm^3 to 6.29 cm^3, and the internal energy of the sample is known to increase by 83.5 J. Use the virial equation of state up to the second coefficient B= -28.7 cm^3 mol^-1 to calculate w, q, and delta H for this change in state."

My reasoning was this. For work = -PDeltaV, I derived that Work = -RT(1+ B/delta Vm) since the virial equation of state is PVm= RT(1 + B/Vm). Once I got work, I could calculate q since delta U (given) is equal to Q + W. For delta H, since H = U + PV, I would have delta H = delta U + RT(1 + B/delta Vm). If you could tell me if my reasoning is correct, that would be great. If it is not, please let me know what I did wrong and how to correct that.

The reasoning seems ok to me, but I don't think the work expression you found is correct. You should integrate -pdV between V₁ and V₂ writing p using the virial equation of state.

Also, for the value of R in that equation, would I use 0.08206 or 8.314? Since I'm in cm^3 and they're both in units dm^3 or m^3.

Unless you want the answers in L·atm, which is not the most common energy unit, you'd better do a couple of conversions and use m³ to get the energy values in J (and R=8.314 J·mol^-1·K^-1).

[orgo814]

I considered integrating which would give me work = -nrt ln vf/vI but I thought that was only for a perfect gas

[MrTeo]

I considered integrating which would give me work = -nrt ln vf/vI but I thought that was only for a perfect gas

You should integrate -pdV between V₁ and V₂ writing p using the virial equation of state.

[orgo814]

How would I go about the integration then since I was wrong

[orgo814]

I think I figured it out but let me do the calculations in a bit when i have time and then ill put it on here for you to check

[orgo814]

OK. So, here are my thoughts.

I derived that w = -RT (ln Vm2/Vm1 - B (1/Vm2- 1/Vm1). I checked online and a source tells me it should be W = -NRT (ln Vm2/Vm1- B (1/Vm2- 1/Vm1). So, I'm not sure why we would use NRT and not RT especially since we're dealing with molar volumes. Insight on that would be appreciated..

Also, for the H = U + PV, when I told you how I was planning on calculating it, does it matter that the PV in that equation isn't based off molar volume (just regular volume) unlike the virial equation. I sort of just assumed it wouldn't matter and that's how I went through with it.. my reasoning on how I did it explained in the question.. but I just want to make sure it's not a big deal.

[MrTeo]

OK. So, here are my thoughts.

I derived that w = -RT (ln Vm2/Vm1 - B (1/Vm2- 1/Vm1). I checked online and a source tells me it should be W = -NRT (ln Vm2/Vm1- B (1/Vm2- 1/Vm1). So, I'm not sure why we would use NRT and not RT especially since we're dealing with molar volumes. Insight on that would be appreciated..

Now the work expression is OK. Since the virial state equation is written using molar volumes the fact that you find the work for one mole makes sense: if you repeat the same process n times you will have

[tex] w_{total} = nw_{molar} = -nRT\left(\ln{\frac{V_{m,2}}{V_{m,1}}}-B\left(\frac{1}{V_{m,2}}-\frac{1}{V_{m,1}}\right)\right)[/tex]

which is exactly what you found as a general expression (work of n moles).

Also, for the H = U + PV, when I told you how I was planning on calculating it, does it matter that the PV in that equation isn't based off molar volume (just regular volume) unlike the virial equation. I sort of just assumed it wouldn't matter and that's how I went through with it.. my reasoning on how I did it explained in the question.. but I just want to make sure it's not a big deal.

Knowing the virial equation you can write ∆(pV_m) as ∆(RT(1+B/V_m)) which you can calculate with the data you have. But ∆(pV)=n∆(pV_m) as volume is the only extensive property. So...