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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? $$ \begin{array}{llllll} \hline \begin{array}{l} \text { Pressure } \\ \text { (atm) } \end{array} & 1.00 & 0.666 & 0.500 & 0.333 & 0.250 \\ \text { Density } & & & & & \\ \text { (g/L) } & 2.3074 & 1.5263 & 1.1401 & 0.7571 & 0.5660 \\ \hline \end{array} $$

Short Answer

Expert verified
The molar mass M of the unknown gas can be calculated by first creating a table of density (d) divided by pressure (P) for each given data point. Plot the d/P values against P, and fit either a linear or non-linear line (or curve) through the data points. The slope of this line corresponds to M/RT, from which M can be found since R (ideal gas constant) and T (temperature) are known. The reason d/P is not a constant function of pressure is that the Ideal Gas Law assumes no intermolecular forces between gas molecules, while in reality, these forces do exist and can affect gas behavior as pressure changes.

Step by step solution

01

Calculate d/P for each data point

Create a new table where there is a new column of the values d/P. This table will be used to find a relationship between d/P and P and thus find the constant M.
02

Plot the graph of d/P versus P

Using the data points calculated in Step 1, plot a graph of d/P on the y-axis and P on the x-axis. The graph should have a recognizable pattern or trend that can be used to fit a line or curve.
03

Fit a line (or curve) through the data points

Carefully fit a line, if the graph is linear, or a curve, if the graph is nonlinear, through the data points by applying curve fitting techniques or using an available software (e.g., MS Excel, Origin, MATLAB). Find the equation of the line (or curve) you just fit.
04

Determine the value of M

The slope of the fitted line in Step 3 would provide the value of M/RT. Since R (the ideal gas constant) and T (temperature, given as 0°C or 273.15 K) are known, M can be calculated as follows: \( M = \frac{slope \times R \times T}{1} \) Once M is calculated, the molar mass of the gas can be determined.
05

Answer part (b) of the question

The question asks why d/P is not a constant function of pressure. The reason for this deviation is that the Ideal Gas Law assumes that the gas molecules do not interact with each other, which is not always true. In reality, gases do have intermolecular forces acting between them, so d/P can change as the pressure increases or decreases due to changes in these forces.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Molar Mass Determination
Determining the molar mass of a gas is an important aspect in understanding its properties and behavior. To do this, we primarily utilize the relationship between pressure, density, and temperature of a gas. By plotting the ratio of density to pressure (\( \frac{d}{P} \)) against pressure, we can determine a constant that represents the molar mass, given certain conditions are met.

Typically, this approach involves plotting experimental data and fitting a curve or line to the plot. The slope of this line, under ideal conditions, is directly related to the gas's molar mass. Given this relationship, by knowing the gas constant \( R \) and the temperature \( T \), we are able to produce a calculation as follows: \( M = \frac{slope \times R \times T}{1} \).

This equation provides a bridge between experimental measurements and theoretical predictions, allowing us to anchor the character of the gas within the broader frameworks of chemistry and physics.
Ideal Gas Law
The Ideal Gas Law is a crucial equation in understanding the behavior of gases under various conditions. It is often expressed as \( PV = nRT \), where \( P \) represents pressure, \( V \) volume, \( n \) the number of moles, \( R \) the universal gas constant, and \( T \) the temperature in Kelvin.

It assumes that a gas consists of many tiny particles that are in constant motion and do not interact with each other. While real gases sometimes deviate from this behavior, the Ideal Gas Law provides an excellent approximation under a wide range of conditions.

This law is fundamental in calculating crucial attributes such as molar mass, using observed properties such as pressure and volume. In situations where deviations from ideal behavior occur, adjustments may be made, but the Ideal Gas Law remains a core principle in gas chemistry, offering foundational insights into the relational dynamics between pressure, volume, temperature, and mole quantity.
Pressure and Density Relationship
The relationship between pressure and density is pivotal in understanding gas behavior. According to the Ideal Gas Law, when temperature and the number of moles are held constant, an increase in pressure on a gas generally results in an increase in density. This is because density (\( d \)) is mass per unit volume, and as per the Ideal Gas Law, any increase in pressure typically compresses the gas, reducing volume and thus increasing density.

Plotting the ratio \( \frac{d}{P} \) versus pressure provides valuable insights into whether a gas follows ideal behavior or exhibits deviations. Ideally, this ratio should remain constant as pressure changes. However, real gases might show a varying ratio due to intermolecular forces or other interactions between gas molecules not accounted for in the Ideal Gas Law.

Understanding this relationship is essential in processes such as calculating molar mass or predicting how a gas will behave under different conditions. It allows students and scientists alike to bridge theoretical concepts with experimental data, deepening the understanding of the interactions at play.
Intermolecular Forces in Gases
Intermolecular forces are the forces of attraction or repulsion between molecules. In gases, these forces are generally weaker than those present in liquids or solids but can still impact gas behavior significantly. As pressure increases and molecules are pushed closer together, these forces become more apparent, leading to deviations from ideal gas behavior.

These deviations can cause the density-pressure ratio (\( \frac{d}{P} \)) to fluctuate, rather than remain constant as predicted by the Ideal Gas Law. At higher pressures, gases can experience forces such as Van der Waals forces, which cause them to deviate from the predictions of the Ideal Gas Law.

Understanding these interactions is crucial, especially when considering gases under conditions where high pressure or low temperature might cause significant deviations. This knowledge helps in accurately modeling and predicting gas behavior in real-world scenarios, such as in industrial applications or atmospheric studies.

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Most popular questions from this chapter

Consider a mixture of two gases, \(A\) and \(B\), confined in a closed vessel. A quantity of a third gas, \(\mathrm{C}\), is added to the same vessel at the same temperature. How does the addition of gas \(\mathrm{C}\) affect the following: (a) the partial pressure of gas \(A,(b)\) the total pressure in the vessel, (c) the mole fraction of gas B?

(a) How is the law of combining volumes explained by Avogadro's hypothesis? (b) Consider a 1.0-L flask containing neon gas and a 1.5-L flask containing xenon gas. Both gases are at the same pressure and temperature. According to Avogadro's law, what can be said about the ratio of the number of atoms in the two flasks?

Assume that an exhaled breath of air consists of \(74.8 \% \mathrm{~N}_{2}\), \(15.3 \% \mathrm{O}_{2}, 3.7 \% \mathrm{CO}_{2}\), and \(6.2 \%\) water vapor. (a) If the total pressure of the gases is \(0.980 \mathrm{~atm}\), calculate the partial pressure of each component of the mixture. (b) If the volume of the exhaled gas is \(455 \mathrm{~mL}\) and its temperature is \(37^{\circ} \mathrm{C}\), calculate the number of moles of \(\mathrm{CO}_{2}\) exhaled. (c) How many grams of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) would need to be metabolized to produce this quantity of \(\mathrm{CO}_{2}\) ? (The chemical reaction is the same as that for combustion of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). See Section 3.2.)

Based on their respective van der Waals constants (Table 10.3), is Ar or \(\mathrm{CO}_{2}\) expected to behave more nearly like an ideal gas at high pressures? Explain.

Which of the following statements best explains why a closed balloon filled with helium gas rises in air? (a) Helium is a monatomic gas, whereas nearly all the molecules that make up air, such as nitrogen and oxygen, are diatomic. (b) The average speed of helium atoms is higher than the average speed of air molecules, and the higher speed of collisions with the balloon walls propels the balloon upward. (c) Because the helium atoms are of lower mass than the average air molecule, the helium gas is less dense than air. The balloon thus weighs less than the air displaced by its volume. (d) Because helium has a lower molar mass than the average air molecule, the helium atoms are in faster motion. This means that the temperature of the helium is higher than the air temperature. Hot gases tend to rise.

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