I was wondering whether the compressibility factor of a real gas (given by $Z = V_{\mathrm{real}}/V_{\mathrm{ideal}}$) is supposed to be measured while keeping pressure constant?

I was attempting to answer the question below and got stumped at part (iii) because I had trouble finding out which was the dominating force being that when I tried calculating the compression factor I got a value of 1 (because $V_{\mathrm{real}}$ and $V_{\mathrm{ideal}}$ are the same, but the pressures are significantly different).

3.0 mol $\ce{CO2}$ behaving as a van der Waals gas, when it is confined under the following conditions: at 750 K in $\pu{150 cm^3}$. For $\ce{CO2}$ van der Waals coefficients, a: $\pu{3.610 atm L^{2} mol^{-2}}$, b: $\pu{4.29 \times 10^{-2} Lmol^{-1}}$. The universal gas constant R is $\pu{8.206 * 10^{-2} L atm K^{-1}mol^{-1}}$

i) Calculate the gas pressure (in atm), [5]

ii) What is the molar volume of the gas (in L/mol), [1]

iii) What is the pressure when it is ideal gas and use it to state the dominating force in the above gas?

  • $\begingroup$ Part (iii) is asking which of the van der Waals factors dominates. $\endgroup$ – MaxW Feb 21 '20 at 10:44
  • $\begingroup$ It's compressibility factor not compression factor... $\endgroup$ – Zenix Feb 21 '20 at 11:12
  • $\begingroup$ Please show us what you did so far,, in detail. $\endgroup$ – Chet Miller Feb 21 '20 at 13:47
  • $\begingroup$ You don't need to know the compressibility factor to solve this problem. But, in any event, its definition is $$z=\frac{PV}{nRT}$$ $\endgroup$ – Chet Miller Feb 21 '20 at 14:16

From the given data, the molar volume $V_\mathrm m$ is $$V_\mathrm m=\frac{150\ \mathrm{cm^3}}{3\ \mathrm{mol}}=50\ \mathrm{cm^3/mol}=0.05\ \mathrm{L/mol}$$

In terms of molar volume, the van der Waals equation reads: $$\left(p+\frac{a}{V_\mathrm m^2}\right)(V_\mathrm m-b)=RT$$

So, solving for the pressure gives:
$$p=\frac{RT}{V_\mathrm m-b}-\frac{a}{V_\mathrm m^2}$$

So, to get the pressure you don't need to know the compressibility factor. Which of these two terms dominates numerically in determining the pressure?




cancelling $n$ on both sides

write $pV$ as $znRT$ where $z$ is compressiblity factor
$$(z-1)RT=pb-a\frac nV+ab\left(\frac nV\right)^2$$

it is behaving like ideal gas hence $z=1$ and right hand side is $0$

$$pb-a\frac nV+ab\left(\frac nV\right)^2=0$$

substitute the values and get the result

  • 3
    $\begingroup$ Welcome on the ChemSE! We have quite good latex support here, I suggested an improvement to your post. After accepting the suggestion, you could end the work. And, please formulate in clear, round sentences. $\endgroup$ – peterh Feb 21 '20 at 15:01
  • 1
    $\begingroup$ See math.meta.stackexchange.com/questions/5020/… for an easy guide to formatting equations. $\endgroup$ – porphyrin Feb 21 '20 at 16:48

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