Question 5.8: In Example Problem 5.1, we considered a CH4 storage tank wit......
In Example Problem 5.1, we considered a CH_4 storage tank with a volume of 49.0 L. When empty, the tank has a mass of 55.85 kg, and when filled, its mass is 62.07 kg. Calculate the pressure of CH_4 in the tank at an ambient temperature of 21°C using both the ideal gas equation and the van der Waals equation. What is the percentage correction achieved by using the more realistic van der Waals equation?
Strategy We will use two different models to describe the same gas, and each model will be represented by its own equation. So we will do two independent calculations, solving for P in each case. We can find the mass of CH_4 from the given data and then convert that mass into the number of moles by using the molar mass. And the van der Waals constants for CH_4 can be found in Table 5.2. To calculate the percentage difference we will divide the difference of the two numbers by the value for the ideal gas case.
| Table 5.2 | ||
| Van der Waals constants for several common gases | ||
| Gas | a (atm L² mol^{−2}) |
b (L mol^{−1}) |
| Ammonia, NH_3 | 4.170 | 0.03707 |
| Argon, Ar | 1.345 | 0.03219 |
| Carbon dioxide, CO_2 | 3.592 | 0.04267 |
| Helium, He | 0.034 | 0.0237 |
| Hydrogen, H_2 | 0.2444 | 0.02661 |
| Hydrogen fluoride, HF | 9.433 | 0.0739 |
| Methane, CH_4 | 2.253 | 0.04278 |
| Nitrogen, N_2 | 1.390 | 0.03913 |
| Oxygen, O_2 | 1.360 | 0.03183 |
| Sulfur dioxide, SO_2 | 6.714 | 0.05636 |
| Water, H_2O | 5.464 | 0.03049 |
Mass of CH_4 = mass of full cylinder − mass of empty cylinder
= 62.07 kg − 55.85 kg = 6.22 kg = 6220 g
Moles of gas =6220\mathrm{~g~CH}_{4}\times{\frac{1\mathrm{~mol~CH}_{4}}{16.04\mathrm{~g~CH}_{4}}}=388\mathrm{~mol~CH}_{4}
Ideal Gas:
P={\frac{n R T}{V}}={\frac{(388\;\mathrm{mol})(0.08206\;\mathrm{L\;atm\;mol}^{-1}\;\mathrm{K}^{-1})(294\;\mathrm{K})}{49.0\;\mathrm{L}}}=191\;\mathrm{atm}
van der Waals:
a=2.253\ \mathrm{L^{2}}\operatorname{atm}\operatorname{mol}^{-2},b=0.04278\ \mathrm{L\ mol^{-1}}
P={\frac{n R T}{V-n b}}-{\frac{n^{2}a}{V^{2}}}
=\frac{(388\;\mathrm{mol})(0.08206\;\mathrm{L~atm~mol}^{-1}\;\mathrm{K}^{-1})(294\;\mathrm{K})}{49.0\;\mathrm{L}-(388\;\mathrm{mol})(0.0428~\mathrm{L}\;\mathrm{mol}^{-1})}
-\;\frac{(388\;\mathrm{mol})^{2}(2.25\;\mathrm{L}^{2}\;\mathrm{atm\;mol}^{-2})}{(49.0\;\mathrm{L})^{2}}
= 148 atm
Percentage correction:
\frac{191-148}{191}\times 100% = 23%
Discussion The correction here is quite significant. If you look at the calculated pressures, you should realize that they are quite high. Recall that the ideal gas law works best at low pressures and high temperatures. In this example, the pressure is sufficiently high that the ideal gas model is a rather poor description of the actual behavior of the gas. Still, the simplicity of the ideal gas law makes it a good starting point for many calculations. In a practical application, it would be crucial for an engineer to recognize the conditions under which the assumption of ideal gas behavior might not be reasonable.
Check Your Understanding Under less extreme conditions, the corrections for nonideal behavior are smaller. Calculate the pressure of 0.500 mol of methane occupying 15.0 L at 25.7°C using both the ideal gas law and the van der Waals equation. Compare the percentage correction with that calculated in the example above.