Problem 103 - Chemistry: The Central Science

Question

The density of a gas of unknown molar mass was measured as a function of pressure at \(0{ }^{\circ} \mathrm{C}\), as in the table below. (a) Determine a precise molar mass for the gas. Hint: Graph \(d / P\) versus \(P\). (b) Why is \(d / P\) not a constant as a function of pressure? $$ \begin{array}{llllll} \hline \text { Pressure } & & & & & \\ \begin{array}{l} \text { (atm) } \end{array} & 1.00 & 0.666 & 0.500 & 0.333 & 0.250 \\ \text { Density } & & & & & \\ \begin{array}{l} \text { (g/L) } \end{array} & 2.3074 & 1.5263 & 1.1401 & 0.7571 & 0.5660 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The average molar mass for the unknown gas is approximately \(M \approx 51.44~g/mol\). The ratio of density to pressure (\(d/P\)) is not constant due to deviations from the ideal gas law, which assumes gas particles are infinitely small and don't interact with each other. Real gases have some volume and intermolecular forces, which affects their behavior especially at high pressures and low temperatures. Despite these deviations, the average molar mass provides an accurate value.

Step 1: Plot the given data

Now, let's plot the provided density by pressure (\(d/P\)) against pressure (\(P\)): \(P\) (atm)|\(d\) (g/L)|\(d/P\) (g/(L·atm)) ---|---|--- 1.00|2.3074|2.3074 0.666|1.5263|2.2913 0.500|1.1401|2.2802 0.333|0.7571|2.2736 0.250|0.5660|2.2640 From the plotted data, we can observe that the ratio \(d/P\) is nearly constant for different values of \(P\).

Step 2: Use the ideal gas law to find molar mass

The ideal gas law states: \(PV=nRT\), where: P = pressure, V = volume, n = number of moles, R = the universal gas constant, T = temperature. We can rewrite the equation in terms of molar mass (M), density (d), and the mass of the gas (m): \(d=\frac{m}{V}=\frac{nM}{V}\). So, the equation becomes: \(P=\frac{nM}{d}RT\), and after isolating M, we get: \(M = \frac{dRT}{P}\). Now, we can calculate the molar mass for the gas using the given data and average the results: Temperature: \(T = 0^{\circ}C = 273.15K\) Gas constant: \(R = 0.0821\frac{L·atm}{mol·K}\) Molar mass for different \(P\): \(M_1 = \frac{2.3074 \cdot 0.0821 \cdot 273.15}{1.00} = 51.46~g/mol\) \(M_2 = \frac{1.5263 \cdot 0.0821 \cdot 273.15}{0.666} = 51.45~g/mol\) \(M_3 = \frac{1.1401 \cdot 0.0821 \cdot 273.15}{0.500} = 51.44~g/mol\) \(M_4 = \frac{0.7571 \cdot 0.0821 \cdot 273.15}{0.333} = 51.44~g/mol\) \(M_5 = \frac{0.5660 \cdot 0.0821 \cdot 273.15}{0.250} = 51.42~g/mol\) So, the average molar mass for the gas is: \(M \approx 51.44~g/mol\).

Step 3: Explain the non-constant ratio of density to pressure

The ideal gas law assumes that the gas particles are infinitely small and don't interact with each other. However, real gases do have some volume and intermolecular forces, which can affect their behavior. These deviations are more significant at high pressures and low temperatures, where the particles are closer together. In our case, the ratio of density to pressure (\(d/P\)) is not constant due to these deviations from the ideal gas law. However, since the deviations are small, the average molar mass we calculated should still provide an accurate value.

Key Concepts

Molar Mass Calculation

Calculating the molar mass of a gas using the ideal gas law is a straightforward process once you understand the relationship between the different variables involved. The ideal gas law equation, \(PV=nRT\), can be modified to find the molar mass \(M\). Here, \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the universal gas constant, and \(T\) is the temperature.

For molar mass calculation, the equation can be transformed to \(M = \frac{dRT}{P}\), where \(d\) is the density \(\left( \frac{m}{V} \right)\). This formula allows us to determine the molar mass \(M\) if the density of the gas and the pressure are known. In the exercise, we used different combinations of \(d\) and \(P\) along with the constants \(R = 0.0821 \frac{L\cdot atm}{mol\cdot K}\) and \(T = 273.15\, K\) to calculate consistent molar masses, leading to an average molar mass of approximately 51.44 g/mol for the gas.

Density and Pressure Relationship

The relationship between the density and pressure of a gas can be observed through the ratio \(d/P\). According to the ideal gas law, if temperature remains constant, density should be directly proportional to pressure. In mathematical terms, \(d/P\) should ideally be constant. However, in real-world experiments, this is rarely perfectly maintained.

When plotting the ratio \(d/P\) against pressure \(P\), one might expect a flat line indicating that the ratio remains steady as pressure changes. But in practice, small variations are often observed. These can stem from experimental errors or more complex interactions in the real gas, leading to non-constant measurements of \(d/P\) under varying pressures. Despite these variations, if the deviations are minor, this relationship can still be used to determine important properties, such as the molar mass, with reasonable accuracy.

Real Gas Deviations

Gases do not always behave as outlined by the ideal gas law, owing primarily to molecular interactions and the finite size of gas molecules. The ideal gas law assumes no interaction between gas molecules and that molecules have negligible volume. However, at high pressures and low temperatures, these assumptions do not hold.

Real gases show deviations because molecules experience attractive and repulsive forces and occupy a significant volume relative to the container. These variations cause the properties of the gas to diverge from those predicted by the ideal gas law, such as when observing changes in \(d/P\) over different pressures.

• At high pressures, molecules are pushed closer together, intensifying intermolecular forces.
• At lower pressures, gases tend to behave more ideally, as particles are farther apart, minimizing interactions.
• Temperature also plays a crucial role; lower temperatures increase deviations as kinetic energy is reduced, making molecular attractions relatively stronger.

Understanding these deviations helps improve calculations in real-world scenarios, ensuring better precision in fields that rely heavily on gas behavior.