Calculating DeltaH for Van der Waals Gas

[xmflea]

A sample consisting of 1.00 mol of a van der Waals gas is compressed from 20.0 L to 10.0 L at 300 K. In the process, 20.2 kJ of work is done on the gas. Given that u = {(2a/RT)-b}/Cp,m with Cp,m = 38.4 J K-1mol-1, a = 3.60 L2 atm mol-2, and b = 0.44 L mol-1, calculate deltaH for the process.

Equations: deltaH = nCpdT

Attempt:

deltaU = w

1. H = U + PV DeltaH = q

2. dH = (partialH/partialT)sub(p)DT + (partialH/partialP)sub(T)DP

3. Cp = q/dT = (partialH/partialT)sub(p)

4. deltaH = CpdT

5. so dH = deltaH + (partialH/partialP)sub(T)DP

and then dH = deltaH - CpudP

gotta figure out what deltaH is, but to do that i got to find dH, but that just leads me back to equation 2. and then I am just going round in circles. need help. thanks.

The Attempt at a Solution

 

[Nino MMP]

:deltaH = nCpdT = (1.00 mol)(38.4 J K-1mol-1)(300 K) = 11,520 J dH = deltaH - CpudP = 11,520 J - (38.4 J K-1mol-1)(20.2 kJ/10.0 L) = 4,844 J Therefore, deltaH = 4,844 J

 

[Blueshift5]

:

To calculate deltaH, we need to find the change in enthalpy (dH) for the process. This can be done by using the ideal gas law and the given Van der Waals equation:

1. PV = nRT

2. (P + a(n/V)^2)(V - nb) = nRT

3. P = (nRT/V) - a(n/V)^2 - nb

4. dH = (partialH/partialT)sub(p)dT + (partialH/partialP)sub(T)dP

5. dH = Cp,dT + (V - nb)(partialP/partialT)sub(V)dT + (nRT/V)(partialV/partialT)dT - (nRT/V)(partialb/partialT)dT

6. dH = Cp,dT + (V - nb)(partialP/partialT)sub(V)dT + (nRT/V)(partialV/partialT)dT - (nRT/V)(partialb/partialT)dT

7. dH = Cp,dT + (V - nb)(nR/V)dT + (nRT/V)(-b/V)dT - (nRT/V)(0)dT

8. dH = Cp,dT + (nR/V)(V - nb - b)dT

9. dH = Cp,dT + (nR/V)(V - 2b)dT

10. dH = Cp,dT + (nR/V)(V - 2b)dT

11. dH = Cp,dT + (nR/V)(V - 2b)(dT/dT)

12. dH = Cp,dT + (nR/V)(V - 2b)

13. dH = Cp,dT + (nR/V)(V - 2b)

14. dH = Cp,dT + (nR/V)(V - 2b)

15. dH = Cp,dT + (nR/V)(V - 2b)

16. dH = Cp,dT + (nR/V)(V - 2b)

17. dH = Cp,dT + (nR/V)(V - 2b)

18. dH = Cp,dT + (nR/V)(V - 2b)

19. dH = Cp,dT + (nR