Problem 96 - Chemistry The Central Science In Si Units

Question

The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through it at a known temperature and pressure. In an experiment, \(8.00 \mathrm{~L}\) of argon gas is passed through \(11.7872 \mathrm{~g}\) of liquid hexane \(\mathrm{C}_{6} \mathrm{H}_{14}\) at \(30.0^{\circ} \mathrm{C}\). The mass of the remaining liquid after the experiment is \(4.875 \mathrm{~g}\). Assuming that the gas becomes saturated with hexane vapor and that the total gas volume and temperature remain constant, what is the vapor pressure of hexane in atm?

Step-by-Step Solution

Answer

The vapor pressure of hexane is approximately 0.200 atm.

Step 1: Calculate the mass of hexane evaporated

First, determine how much hexane has been evaporated by subtracting the final mass of hexane from the initial mass. The initial mass is \(11.7872 \text{ g}\) and the final mass is \(4.875 \text{ g}\). Thus, the mass of hexane evaporated is \(11.7872 - 4.875 = 6.9122 \text{ g}\).

Step 2: Calculate moles of hexane evaporated

Convert the mass of evaporated hexane into moles. The molar mass of hexane (\(\mathrm{C}_{6} \mathrm{H}_{14}\)) is approximately \(86.18 \text{ g/mol}\). Therefore, the moles of hexane is \(\frac{6.9122 \text{ g}}{86.18 \text{ g/mol}} \approx 0.0802 \text{ mol}\).

Step 3: Use Ideal Gas Law for Argon

We need to find the pressure using the amount of moles. Start by finding the moles of argon gas. The initial volume of argon is \(8.00 \text{ L}\) at the experiment conditions, \(30.0^{\circ}\text{C} = 303.15 \text{ K}\). The ideal gas law is \(PV = nRT\). Let's assume the pressure \(P\) of argon is initially \(1 \text{ atm}\).Using these, calculate the initial moles of argon: \(n = \frac{PV}{RT}\), where \(R = 0.0821 \frac{L\cdot atm}{mol\cdot K}\). Thus, \(n = \frac{(1 \text{ atm}) \times (8 \text{ L})}{0.0821 \frac{L\cdot atm}{mol\cdot K} \times 303.15 \text{ K}} \approx 0.321 \text{ moles argon}\).

Step 4: Calculate total moles of gas

Since hexane has evaporated and saturated the gas, total moles of gas at the end includes both hexane and argon. Total moles \( = 0.321 \text{ moles argon} + 0.0802 \text{ moles hexane} \approx 0.4012 \text{ moles}\).

Step 5: Apply Dalton's Law of Partial Pressures

According to Dalton's Law, the total pressure of a gas mixture is the sum of the partial pressures of each individual component. Thus, for hexane vapor:\[P_{\text{total}} = P_{\text{argon}} + P_{\text{hexane}}\]However, since argon volume remains constant, the partial pressure of hexane is given by the change in mole fraction:\[P_{\text{hexane}} = P_{\text{total}} \cdot \frac{n_{\text{hexane}}}{n_{\text{total}}}\]Assume the experiment is designed such that the total pressure remains at the initial 1 atm (meaning the equipment can account for this or is readjusted automatically).Thus, \[P_{\text{hexane}} = 1 \text{ atm} \cdot \frac{0.0802}{0.4012} \approx 0.200 \text{ atm}\].

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental principle used to relate the pressure, volume, temperature, and number of moles of an ideal gas. The formula is given by: \( PV = nRT \), where:

• \( P \) is the pressure of the gas in atmospheres (atm),
• \( V \) is the volume of the gas in liters (L),
• \( n \) is the number of moles of the gas,
• \( R \) is the ideal gas constant with a value of \( 0.0821 \ \frac{L\cdot atm}{mol\cdot K} \),
• \( T \) is the temperature in Kelvin (K).

This law is a handy tool for solving problems involving gases, like figuring out unknown quantities when others are known. For example, you can determine the number of moles of a gas if its volume, pressure, and temperature are known, as seen in the exercise where moles of argon were calculated. Understanding this equation is crucial for calculating gas properties under different conditions.

Dalton's Law of Partial Pressures

Dalton’s Law of Partial Pressures tells us that the total pressure in a mixture of gases is equal to the sum of the partial pressures of the individual gases in the mixture. Mathematically, it can be expressed as: \( P_{\text{total}} = P_{1} + P_{2} + P_{3} + \ldots \).
In this exercise, argon gas and hexane vapor together make the total pressure. This principle helps us understand how each gas behaves as though it were alone in the container. The partial pressure of hexane in the gas mixture is computed by using the mole fraction of hexane:
\[ P_{\text{hexane}} = P_{\text{total}} \times \frac{n_{\text{hexane}}}{n_{\text{total}}} \]This calculation shows how much of the total pressure is due to the hexane vapor alone. It’s a thoughtful way to break down how pressure contributions are made by components in a gas mixture.

Volatile Liquids

Volatile liquids are substances that readily vaporize at room temperature. In simpler terms, these are liquids that can easily change into gas. Their ability to vaporize plays a key role in determining their vapor pressure, which is the pressure exerted by their vapor when the liquid and vapor are in equilibrium.
Hexane is one such volatile liquid and its properties are explored in the exercise. When a volatile liquid is placed in a closed system, its vapor pressure can be determined by measuring how much liquid has evaporated. This vaporization proceeds until equilibrium is achieved. Understanding volatility is crucial for applications like predicting evaporation rates and studying the liquid’s behavior in different conditions.

Molar Mass Calculation

The molar mass is the mass of one mole of a substance and is often used to convert between the mass of a substance and the number of moles. This conversion is vital for determining how many molecules of a substance there are in a given mass.
To find the moles of a substance, you use the formula:

• \( n = \frac{\text{mass}}{\text{molar mass}} \)

For hexane, with a molar mass of approximately 86.18 g/mol, this calculation allows us to find how many moles are present in a certain mass, which is key to later calculations involving gas laws. Molar mass is an essential concept because it links the macroscopic properties of a substance to its microscopic molecular scale.