Problem 70
Question
A 1.072 g sample of \(\mathrm{He}(\mathrm{g})\) is found to occupy a volume of 8.446 L when collected over hexane at \(25.0^{\circ} \mathrm{C}\) and \(738.6 \mathrm{mmHg}\) barometric pressure. Use these data to determine the vapor pressure of hexane at \(25^{\circ} \mathrm{C}\).
Step-by-Step Solution
Verified Answer
The vapor pressure of hexane at 25.0°C is \(0.1\) atm. Remember, results may need onward conversion if other units are required.
1Step 1: Convert units and conditions
First, convert all the units to their standard counterparts. Mass of helium (He) = 1.072g. The molar mass of helium is 4.00 g/mol, so the moles of He is given by 1.072 g / 4.00 g/mol = 0.268 mol. The volume of He is given as 8.446 L. Temperature is given in Celsius, so it must be converted into Kelvin by adding 273.15. So, T = 25.0 °C + 273.15 = 298.15 K. The barometric pressure is given in mmHg; to convert it into atmospheres, divide by 760. So, P = 738.6 mmHg / 760 = 0.972 atm.
2Step 2: Apply the ideal gas law
We assume the total pressure is the sum of the pressure of He and the vapor pressure of hexane by Dalton's law of partial pressures. We first need to find the pressure contributed by helium using the ideal gas law. Plugging in the values into PV = nRT, we get (P_He)(8.446 L) = (0.268 mol)(0.0821 L·atm/(K·mol))(298.15 K). Solving for P_He, we discover it equals to 0.872 atm.
3Step 3: Calculate the vapor pressure of hexane
According to Dalton's law, the total pressure is the sum of the pressure of He and the vapor pressure of hexane (P_hexane). So, P_hexane + P_He = P_total. Rearrange to find P_hexane = P_total - P_He. Plugging in the values we have, P_hexane = 0.972 atm - 0.872 atm = 0.1 atm. Note that the pressure units are in atm, so if required in another unit, appropriate conversion is necessary.
Key Concepts
Vapor PressureDalton's Law of Partial PressuresGas Collection over Water
Vapor Pressure
Vapor pressure is a fundamental concept in chemistry that relates to the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases at a given temperature. This means that in a closed system, the vapor of a liquid will exert a pressure as molecules escape from the surface of the liquid into the gaseous phase.
The vapor pressure increases with temperature because more molecules have enough energy to escape the liquid's surface as the temperature rises.
Understanding vapor pressure is important when dealing with gases collected over a liquid, such as in this exercise with helium collected over hexane.
Additionally, vapor pressure is a key property influencing processes like boiling and evaporation. It signifies the point at which the liquid begins to boil when its vapor pressure equals the surrounding atmospheric pressure.
Here, you need to understand that when the helium is collected over hexane, both gases contribute to the total pressure measured but each does so based on their own vapor pressures.
Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is an essential principle when dealing with mixtures of gases. This law states that in a mixture of non-reacting gases, the total pressure of the mixture is equal to the sum of the individual partial pressures of the gases present. The partial pressure is the pressure a gas would exhibit if it occupied the entire volume of the mixture at the same temperature. For any component of a gas mixture, its partial pressure can be calculated as:\[ P_{component} = \frac{nRT}{V} \]Where:- \( n \) is the number of moles of the gas,- \( R \) is the universal gas constant,- \( T \) is temperature in Kelvin,- \( V \) is the volume.In solving the problem, Dalton's law helps us understand how to calculate the individual partial pressures of helium and hexane's vapor and how these contribute to the total pressure of the system.
Gas Collection over Water
The concept of gas collection over a liquid or water is typical in many laboratory settings. It involves capturing a gas that evolves during a reaction or is released and measures the total pressure in the system, which includes both the gas of interest and the vapor of the liquid over which it is collected.
Often, experiments involve collecting gases over water, where the vapor pressure of water must be taken into account when calculating the amount of gas. The principle, however, applies to any liquid, including hexane here.
The procedure involves:
- Recognizing that the total pressure in the collection vessel is not just from the gas being collected but also includes the vapor pressure of the collecting liquid.
- Subtracting the vapor pressure of the liquid from the total pressure yields the pressure of the collected gas. This is critical for accurate measurements.
Other exercises in this chapter
Problem 68
An \(89.3 \mathrm{mL}\) sample of wet \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(21.3^{\circ} \mathrm{C}\) at a barometric pressure of \(756 \m
View solution Problem 69
A sample of \(\mathrm{O}_{2}(\mathrm{g})\) is collected over water at \(24^{\circ} \mathrm{C}\) The volume of gas is 1.16 L. In a subsequent experiment, it is d
View solution Problem 71
At elevated temperatures, solid sodium chlorate \(\left(\mathrm{NaClO}_{3}\right)\) decomposes to produce sodium chloride, \(\mathrm{NaCl},\) and \(\mathrm{O}_{
View solution Problem 72
When solid \(\mathrm{KClO}_{3}\) is heated strongly, it decomposes to form solid potassium chloride, \(\mathrm{KCl}\), and \(\mathrm{O}_{2}\) gas. \(\mathrm{A}\
View solution