Compressibility factor (Z) of a real gas 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?
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?