Question

Somebody claims that oxygen gas at \(160 \mathrm{K}\) and \(3 \mathrm{MPa}\) can be treated as an ideal gas with an error of less than 10 percent. Is this claim valid?

Short answer

Answer: To determine the validity of this claim, follow these steps: 1. Convert the van der Waals constants and pressure to the same unit system. 2. Solve for the molar volume (Vm) using the van der Waals equation. 3. Calculate the compressibility factor (Z) with the obtained molar volume. 4. Compare Z value to the acceptable range (0.9 ≤ Z ≤ 1.1). If the calculated Z value falls within the acceptable range, then the claim is valid, and oxygen gas can be treated as an ideal gas with an error of less than 10 percent. If the value is outside this range, the claim is not valid, and oxygen gas at the given conditions cannot be treated as an ideal gas.

Step-by-step solution

Determine the Compressibility Factor (Z) Formula and Constants

We'll be using the van der Waals equation that expresses the compressibility factor (Z) as: $$ Z = \frac{P \cdot V_m}{R \cdot T} $$ Where P is pressure, \(V_m\) is molar volume, R is the gas constant, and T is temperature. In addition, the van der Waals equation is given by: $$ \left(P + \frac{a}{V_m^2}\right)(V_m - b) = R \cdot T $$ Where a and b are van der Waals constants specific to a given gas. For oxygen, the constants are: $$ a = 1.360 \, L^2 \cdot atm \cdot mol^{-2} \\ b = 0.03183 \, L \cdot mol^{-1} $$ We'll need to convert the units of a to match the units used in the problem: \(L^2 \cdot MPa \cdot mol^{-2}\)

Convert the Units of a and the Given Pressure to the Same Unit System

We need to convert the constant a and the pressure (3 MPa) to the same units. In this case, we'll convert a to \(L^2 \cdot MPa \cdot mol^{-2}\) and the pressure to atm. Conversion factors: 1 MPa = 9.86923 atm 1 atm = 101325 Pa Converting a: $$ a = 1.360 \, L^2 \cdot atm \cdot mol^{-2} \times \frac{1\,MPa}{9.86923\,atm} = 0.1377 \, L^2 \cdot MPa \cdot mol^{-2} $$ Converting pressure: $$ P = 3\,MPa \times 9.86923\,atm/MPa = 29.61\,atm $$

Solve for Molar Volume \(V_m\) Using van der Waals Equation

Now we can use the van der Waals equation to solve for the molar volume (\(V_m\)). First, we'll plug in our values for P, a, b, R, and T: $$ \left(29.61\,atm + \frac{0.1377\,L^2 \cdot MPa \cdot mol^{-2}}{V_m^2}\right)(V_m - 0.03183\,L \cdot mol^{-1}) = 8.314\,J \cdot mol^{-1} \cdot K^{-1} \times 160\,K $$ This is a cubic equation in \(V_m\). Solving this equation by either numerical or graphical methods will give us the molar volume \(V_m\).

Calculate the Compressibility Factor (Z)

Once we have determined the molar volume \(V_m\), we can calculate the compressibility factor Z using the formula: $$ Z = \frac{P \cdot V_m}{R \cdot T} $$ Plug in the values for P, \(V_m\), R, and T, and then calculate Z: $$ Z = \frac{29.61\,atm \cdot V_m}{8.314\,J \cdot mol^{-1} \cdot K^{-1} \times 160\,K} $$

Compare the Compressibility Factor (Z) to the Acceptable Range

The claim states that oxygen gas at the given conditions can be treated as an ideal gas with an error of less than 10 percent. This means that we should compare our calculated Z to 1 to see if it is within 10% of 1. The acceptable range for Z, in this case, would be: $$ 0.9 \leq Z \leq 1.1 $$ If the calculated Z value falls within this range, then oxygen gas can be treated as an ideal gas with an error of less than 10 percent, validating the claim. If the value is outside of this range, then the claim is not valid and oxygen gas at the given conditions cannot be treated as an ideal gas.