Question

A tank contains \(310 \mathrm{~kg}\) of a gaseous mixture of \(70 \%\) ethane and \(30 \%\) nitrogen (molar basis) at \(311 \mathrm{~K}\) and \(170 \mathrm{~atm}\). Determine the volume of the tank, in \(\mathrm{m}^{3}\), using data from the, generalized compressibility chart together with (a) Kay's rule, (b) the ideal solution model. Compare with the measured tank volume of \(1 \mathrm{~m}^{3}\).

Short answer

Using Kay's rule, the volume is approximately 12.38 m³. Using the ideal gas model, it is approximately 15.77 m³. Both are larger than the measured volume of 1 m³.

Step-by-step solution

- Determine the molar mass of the mixture

Calculate the molar mass of each component. Ethane (\text{C}_2\text{H}_6) has a molar mass of 30.07 \text{g/mol} and nitrogen (\text{N}_2) has a molar mass of 28.01 \text{g/mol}. Using the mole fraction, the molar mass of the mixture is: \[ M_{\text{mixture}} = (0.7 \times 30.07) + (0.3 \times 28.01) = 29.33 \text{g/mol} \].

- Determine the number of moles in the tank

Calculate the total number of moles in the tank by dividing the total mass by the molar mass: \[ n = \frac{310 \text{kg}}{29.33 \text{g/mol}} = \frac{310,000 \text{g}}{29.33 \text{g/mol}} \ n \thickapprox 10,571 \text{mol} \].

- Use the compressibility chart with Kay's rule

Determine the pseudocritical properties (pseudocritical temperature and pressure) of the mixture: \[ T_{\text{c, mix}} = (0.7 \times T_{\text{c, ethane}}) + (0.3 \times T_{\text{c, nitrogen}}) \ T_{\text{c, ethane}} = 305.3 \text{K}, \ T_{\text{c, nitrogen}} = 126.2 \text{K}, \ T_{\text{c, mix}} = (0.7 \times 305.3 \text{K}) + (0.3 \times 126.2 \text{K}) = 243.49 \text{K} \]. Similarly for pressure: \[ P_{\text{c, mix}} = (0.7 \times P_{\text{c, ethane}}) + (0.3 \times P_{\text{c, nitrogen}}) \ P_{\text{c, ethane}} = 48.72 \text{atm}, \ P_{\text{c, nitrogen}} = 33.5 \text{atm}, \ P_{\text{c, mix}} = (0.7 \times 48.72 \text{atm}) + (0.3 \times 33.5 \text{atm}) = 44.496 \text{atm} \].

- Calculate reduced temperature and pressure

Determine the reduced temperature and reduced pressure using the actual conditions: \[ T_r = \frac{T}{T_{\text{c, mix}}} = \frac{311 \text{K}}{243.49 \text{K}} = 1.276 \ P_r = \frac{P}{P_{\text{c, mix}}} = \frac{170 \text{atm}}{44.496 \text{atm}} = 3.822 \].

- Find the compressibility factor (Z) from the compressibility chart

Using the compressibility chart, locate the lines for T_r = 1.276 and P_r = 3.822 . Read off the compressibility factor Z, which is approximately \[ Z = 0.7 \].

- Calculate the volume using Kay's rule

Use the real gas equation of state to find the volume: \[ V = \frac{ZnRT}{P} \ V = \frac{0.7 \times 10,571 \text{mol} \times 0.0821 \text{atm·L/mol·K} \times 311 \text{K}}{170 \text{atm}} = 12.38 \text{m}^3 \].

- Ideal gas solution model

Repeat the calculation assuming ideal gas behavior: \[ V_{\text{ideal}} = \frac{nRT}{P} = \frac{10,571 \text{mol} \times 0.0821 \text{atm·L/mol·K} \times 311 \text{K}}{170 \text{atm}} = 15.77 \text{m}^3 \].

- Comparison with measured volume

Compare the calculated volumes with the measured volume of 1 m³. Observations: The volume calculated with Kay's rule is closer to the measured value, but still significantly higher, implying that real-world factors not accounted for by the model might be at play.

Key concepts

molar mass calculation

In chemistry, the molar mass is the mass of one mole of a substance. When you need to determine the molar mass of a gaseous mixture, you take into account the molar masses of each component in the mixture and their respective proportions. For instance, in our exercise, we are dealing with a mixture of ethane (C\text{\(_2\)}H\text{\(_6\)}) and nitrogen (N\text{\(_2\)}).
Ethane has a molar mass of 30.07 g/mol and nitrogen has a molar mass of 28.01 g/mol. To find the average molar mass of the mixture, use the mole fraction of each gas and sum their contributions.

• Mole fraction of ethane is 70% or 0.7, and for nitrogen, it is 30% or 0.3.
• Thus, the molar mass of the gaseous mixture = 0.7 \times 30.07 + 0.3 \times 28.01 = 29.33 g/mol.

This calculation gives us an important value needed for subsequent steps, such as determining the number of moles in the tank.

compressibility chart

The compressibility chart is an essential tool in the study of gases. It helps determine the compressibility factor (Z), which indicates how much a real gas deviates from ideal gas behavior. This factor is crucial when dealing with high-pressure gases.
To use the chart, you need values of reduced temperature and reduced pressure. These are found by dividing the actual temperature (T) and actual pressure (P) by their respective pseudocritical values (T\text{\(_{c}\)} and P\text{\(_{c}\)}).

• From our example, reduced temperature, T\text{\(_{r}\)} = \frac{T}{T\text{\(_{c, mix}\)}} = \frac{311 \text{K}}{243.49 \text{K}} = 1.276
• Reduced pressure, P\text{\(_{r}\)} = \frac{P}{P\text{\(_{c, mix}\)}} = 170 \text{atm} / 44.496 \text{atm} = 3.822.

Using these reduced values, you can locate the corresponding Z value on the chart. In this exercise, Z is approximately 0.7. This Z value is then used in the real gas equation of state to calculate volume.

Kay's rule

Kay's rule provides a method to estimate the pseudocritical properties of a gas mixture. It is particularly useful when precise data for the mixture aren't available. This rule uses the mole fractions and critical properties of each component:

• For critical temperature, T\text{\(_{c, mix}\)} = (0.7 \times 305.3 \text{K}) + (0.3 \times 126.2 \text{K}) = 243.49 \text{K}
• For critical pressure, P\text{\(_{c, mix}\)} = (0.7 \times 48.72 \text{atm}) + (0.3 \times 33.5 \text{atm}) = 44.496 \text{atm}

Once you have these pseudocritical properties, you can use them to calculate reduced temperature and pressure, which are necessary for finding the compressibility factor from the chart.
This approach is often more accurate than using the ideal gas law, especially under non-ideal conditions like high pressure.

ideal gas law

The ideal gas law is a fundamental equation in chemistry and physics. It states that for an ideal gas, the volume (V) is directly proportional to the number of moles (n), the temperature (T), and inversely proportional to the pressure (P).
The equation is given as:
V = \frac{nRT}{P}
where R is the universal gas constant (0.0821 atm·L/mol·K).

• Using this law: V\text{\(_{ideal}\)} = \frac{10,571 \text{mol} \times 0.0821 \text{atm·L/mol·K} \times 311 \text{K}}{170 \text{atm}} = 15.77 \text{m}^3.

Although the ideal gas law provides a straightforward approach to determining gas volume, it can lead to significant errors at high pressures or low temperatures (conditions where gases deviate from ideality).
That's why comparing the volume calculated using ideal gas law with more complex real gas models, like Kay's rule and compressibility charts, provides better insight.