91Ó°ÊÓ

Molar mass is a key concept in chemistry and thermodynamics that helps us connect moles of substances to their mass in a specific unit. When calculating molar mass, we sum up the atomic masses of all atoms present in a molecule. These atomic masses are typically listed on the periodic table. For gases, molar mass is usually expressed in units like lbm/lb-mol or g/mol.

In our exercise, we looked at a mixture consisting of nitrogen ( _{2} ), water ( _{2}O"), and oxygen ( _{2} ). Their respective molar masses were given as 28.02 lbm/lb-mol for nitrogen, 18.02 lbm/lb-mol for water, and 32.00 lbm/lb-mol for oxygen. Knowing these values is crucial for converting between mole and mass fractions in mixtures.

Mass fraction is a way to express how much of a particular substance makes up a given mixture in terms of mass. It tells us what portion of the total mass of the mixture is due to each component.

To find mass fractions, use the formula:\[ w_i = \frac{x_i \times M_i}{\sum{x_i \times M_i}} \]where:

• \(w_i\) is the mass fraction of component \(i\)
• \(x_i\) is the mole fraction of component \(i\)
• \(M_i\) is the molar mass of component \(i\)

For each gas in our mixture, we used their mole fractions (given in the problem statement) and their molar masses to calculate the mass fraction of each component. This conversion is done by weighting each component's molar mass by its mole fraction, then dividing by the total weighted molar mass sum.

In this case:

• Nitrogen (_{2}): Mass fraction \(\approx 0.6607\)
• Water (_{2}}O): Mass fraction \(\approx 0.255\)
• Oxygen (_{2}): Mass fraction \(\approx 0.0843\)

A gas constant is essential when dealing with thermodynamic properties of gases. In mixtures, we use a special form called the mixture gas constant, \(R_m\). This is derived from the universal gas constant \(R_u\) and is recalculated based on the molar mass of the specific mixture we are analyzing.

The formula used is:\[ R_m = \frac{R_u}{\bar{M}} \]Here:

• \(R_u\) is known as the universal gas constant, which is approximately \(1545.349 \, \text{ft}\cdot\text{lbf}/\text{lb-mol}\cdot\text{R}\)
• \(\bar{M}\) is the apparent molar mass of the gas mixture

For our exercise, the apparent molar mass \(\bar{M}\) of the mixture was calculated as \(21.205 \, \text{lbm}/\text{lb-mol}\), leading us to find that the mixture gas constant \(R_m\) is approximately \(72.84 \, \text{ft}\cdot\text{lbf}/\text{lbm}\cdot\text{R}\). This constant helps us connect pressure, volume, and temperature through the ideal gas law.

The ideal gas law is a cornerstone of thermodynamics and relates the pressure, volume, temperature, and quantity of an ideal gas. The law is usually stated as:\[ PV = nRT \]In situations where mass is given instead of moles, it is often rewritten using the mixture gas constant as:\[ PV = mRT \]Here:

• \(P\) is the pressure
• \(V\) is the volume
• \(m\) is the mass
• \(R\) is the specific gas constant for the mixture
• \(T\) is the absolute temperature

In our example, with a given mass of the mixture (10 lbm), we used the specific form of the ideal gas law formula to find the volume. The pressure was converted to 18 psi, and the temperature in Rankine was calculated for use in the equation. By substituting known values into the ideal gas model equation, we get the volume \(V\approx 28688.8 \, \text{ft}^3\). This demonstrates how these combined concepts allow us to predict the behavior of gases under different conditions.