Enthalpy change for a real gas (Carbon dioxide)

  • #1
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Homework Statement


Find ΔH for the isothermal expansion of one mole of CO2 from a pressure of 1 atm to zero at 300 K. The critical point of CO2 is TC = 31 °C and PC = 73 atm. Use the equation for ΔH you previously derived from the Berthelot equation of state. (Answer provided by textbook: ΔH = -10.2 cal)

Homework Equations


ΔH for isothermal processes of real gases, derived in a previous problem from the Berthelot EOS:
[tex]\Delta H = \left( An + \frac{3Bn}{T^2} \right) \Delta P[/tex]
[tex]A = \frac{9RT_C}{128P_C}[/tex]
[tex]B = -\frac{27RT_C^3}{64P_C}[/tex]

The Attempt at a Solution


Calculating A and B from provided data and using R = 0.08205 L atm mol-1 K-1.
A = 0.02404 L mol-1
B = -13341.448 L K2 mol-1
If we got from 1 atm to zero, then ΔP = -1 atm. Plugging all these values into the ΔH equation gives:
[tex]\Delta H = \left( \left(0.02404 \ \frac{L}{mol} \right)(1 \ mol) + \frac{(3)(-13341.448 \ \frac{L \cdot K^2}{mol})(1 \ mol)}{(300 K)^2} \right) (-1 \ atm) = 0.4207 \ L \cdot atm[/tex]
[tex]\Delta H = 0.4207 \ L \cdot atm \ \left(\frac{24.22 \ cal}{1 \ L \cdot atm} \right) = 10.19 \ cal[/tex]
As you can see, my answer is numerically correct, but positive instead of negative as in the textbook. I've checked every part of the problem and found no error, so I don't know what should I change to get the negative value. Is my ΔP correct? I don't think the minus sign in the answer is a typo, since my intuiton is telling me the system should have a net decrease in energy (as it theoretically ended up having zero pressure), so I must be the one making a mistake. Any insight will be greatly appreciated, thanks in advance!
 

Answers and Replies

  • #2
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Check out a pressure - enthalpy diagram for CO2. You will see that the enthalpy increases with decreasing pressure at constant temperature.

So your answer must be right.

Chet
 
  • #3
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Could it be that the book made the same mistake twice? In a following problem, it states that for CO2 at 300 K and 1 atm, [itex]\left(\frac{\partial \hat{H}}{\partial P} \right)_T = -10.2 \ \frac{cal}{mol \cdot atm}[/itex]. Or does this derivative mean that for every pressure increase of 1 atm, enthalpy will decrease 10.2 calories per mole, thus explaining my positive result?
 
  • #4
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4,300
Could it be that the book made the same mistake twice? In a following problem, it states that for CO2 at 300 K and 1 atm, [itex]\left(\frac{\partial \hat{H}}{\partial P} \right)_T = -10.2 \ \frac{cal}{mol \cdot atm}[/itex]. Or does this derivative mean that for every pressure increase of 1 atm, enthalpy will decrease 10.2 calories per mole, thus explaining my positive result?
The latter.

Actually the equation you originally wrote was kind of confusing. It should represent the partial derivative of enthalpy with respect to pressure, but it contains no pressure dependence inside the parenthesis. We know that, at zero pressure, the partial of H with respect to P is zero, because, in that limit, the gas is an ideal gas. So why is there no pressure dependence inside the parenthesis?

Chet
 
  • #5
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This is how I got to the ΔH equation I used. I started with the differential of H:
[tex]dH = \left(\frac{\partial H}{\partial T} \right)_P dT + \left(\frac{\partial H}{\partial P} \right)_T dP[/tex]
Since we are having an isothermal process, the first term of the RHS is zero. The partial of H with respect to P is defined like this:
[tex]\left(\frac{\partial H}{\partial P} \right)_T = V - T \left(\frac{\partial V}{\partial T} \right)_P[/tex]
We will need this definition a little bit later on. We will use the Berthelot EOS for our calculations:
[tex]PV = nRT \left(1+ \frac{9PT_C}{128P_C T} - \frac{27PT_C^3}{64P_C T^3} \right)[/tex]
After we remove parentheses we arrive at this:
[tex]PV = nRT + \left( \frac{9RT_C}{128P_C} \right)nP + \left( - \frac{27RT_C^3}{64P_C} \right) \frac{nP}{T^2}[/tex]
Letting:
[tex]A=\frac{9RT_C}{128P_C} \ \ \ \ \ \ \ \ \ \ B=-\frac{27RT_C^3}{64P_C}[/tex]
A and B are supposed to be constant for each different gas, notice how they are functions of other known constants. We are left with a much simpler equation:
[tex]PV = nRT + AnP + \frac{BnP}{T^2}[/tex]
Now we clear V and take the partial of V wrt T, in order to calculate the partial of H wrt P and place it on dH:
[tex]V = \frac{nRT}{P} + An + \frac{Bn}{T^2}[/tex]
[tex]\left( \frac{\partial V}{\partial T} \right)_P = \frac{nR}{P} - \frac{2Bn}{T^3}[/tex]
[tex]V-T\left( \frac{\partial V}{\partial T} \right)_P = \frac{nRT}{P} + An + \frac{Bn}{T^2} - \frac{nRT}{P} + \frac{2Bn}{T^2}[/tex]
[tex]dH = \left(An + \frac{3Bn}{T^2} \right) dP[/tex]
As you can see, all terms left in the RHS are constant in this problem. Finally, we just integrate to get ΔH:
[tex]\int_{H_1}^{H_2} dH = \int_{P_1}^{P_2} \left(An + \frac{3Bn}{T^2} \right) dP[/tex]
[tex]\Delta H = \left(An + \frac{3Bn}{T^2} \right) \Delta P[/tex]
 
  • #6
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4,300
I've learned something new from what you did and from researching this further. I always thought that the partial derivative of enthalpy with respect to pressure approaches zero for all real gases as the pressure approaches zero. But apparently this is not the case. Examination of graphs of the compressibility factor z as a function of temperature and pressure clearly indicates that at low pressures, the compressibility factor varies linearly with pressure. This means that the partial derivative of enthalpy with respect to pressure approaches a non-zero function of temperature as the pressure approaches zero. Your results for this EOS model illustrate this.

Chet
 
  • #7
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Thanks for all the input! Glad to hear I contributed with something new, this was enlightnening for me too, as I stated in the OP I thought the negative answer of the book was correct since pressure was decreasing. Real gases are really complex, I guess.
 

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