- #1

- 229

- 53

## Homework Statement

Find ΔH for the isothermal expansion of one mole of CO

_{2}from a pressure of 1 atm to zero at 300 K. The critical point of CO

_{2}is T

_{C}= 31 °C and P

_{C}= 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 K

^{2}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!