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Chapter 5: Problem 100

Helium is collected over water at \(25^{\circ} \mathrm{C}\) and 1.00 atm total pressure. What total volume of gas must be collected to obtain 0.586 g helium? (At \(25^{\circ} \mathrm{C}\) the vapor pressure of water is 23.8 torr.)

Short Answer

Expert verified
To obtain 0.586 g of helium at \(25^{\circ} \mathrm{C}\) and 1.00 atm total pressure, the total volume of helium gas collected over water should be approximately 3.90 L.

Step by step solution

01

Convert the given information to appropriate units

Before we begin, let's first convert the given information to the appropriate units for the Ideal Gas Law equation, which is PV = nRT. Temperature = \(25^{\circ} \mathrm{C}\) = (25 + 273.15) K = 298.15 K Total pressure = 1.00 atm (Already in the correct unit) Vapor pressure of water = 23.8 torr. We need to convert this to atm. \(1 \mathrm{atm} = 760 \mathrm{torr}\) Water vapor pressure = \(\frac{23.8 \mathrm{torr}}{760 \mathrm{torr/atm}} = 0.03132 \mathrm{atm}\) Helium mass = 0.586 g Molar mass of helium = 4.00 g/mol
02

Calculate the partial pressure of helium

Since helium is collected over water, we need to find the partial pressure of helium. The total pressure is the sum of the pressure due to helium and the pressure due to water vapor. Total pressure = Helium pressure + Water vapor pressure Helium pressure = Total pressure - Water vapor pressure \(P_{\mathrm{He}} = 1.00 \mathrm{atm} - 0.03132 \mathrm{atm}\) \(P_{\mathrm{He}} = 0.96868 \mathrm{atm}\)
03

Calculate the moles of helium collected

Next, we need to compute the number of moles of helium collected. To do this, we will use its mass and molar mass: Moles of helium = \(\frac{\mathrm{mass}}{\mathrm{molar \, mass}}\) n = \(\frac{0.586 \, \mathrm{g}}{4.00 \mathrm{g/mol}}\) n 鈮� 0.1465 mol
04

Use the Ideal Gas Law to find the volume

Now that we have the partial pressure of helium, the temperature, and the moles of helium, we can use the Ideal Gas Law to find the volume of gas collected. \(PV = nRT\) \(V = \frac{nRT}{P}\) V = \(\frac{(0.1465 \, \mathrm{mol})(0.0821 \, \mathrm{L \cdot atm/mol \cdot K})(298.15 \, \mathrm{K})}{0.96868 \, \mathrm{atm}}\) V 鈮� 3.90 L
05

Write the final answer

To obtain 0.586 g of helium at \(25^{\circ} \mathrm{C}\) and 1.00 atm total pressure, the total volume of helium gas collected over water should be approximately 3.90 L.

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Key Concepts

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

Partial Pressure in Gas Mixtures
In a mixture of gases, each gas contributes to the total pressure. This contribution is called partial pressure. When gases like helium are collected over water, the total pressure includes the pressure from water vapor.
To calculate the partial pressure of helium, we have to subtract the vapor pressure of water from the total pressure. This is because water vapor is part of the gas mixture.
  • Total Pressure = Helium Pressure + Water Vapor Pressure
  • Helium Pressure = Total Pressure - Water Vapor Pressure
Knowing how to calculate partial pressures helps us utilize the Ideal Gas Law effectively. This understanding is important in experiments involving gas collection over liquids.
Understanding Moles of Helium
The amount of helium in a sample can be measured in moles, which relate to the number of atoms or molecules rather than mass. This is crucial in chemical calculations. To find moles, you divide the mass of the gas by its molar mass.
For helium, with a molar mass of 4.00 g/mol, the number of moles can be calculated as follows:
  • Moles of Helium = \( \frac{\text{Mass of helium}}{\text{Molar mass}} \)
  • \( n = \frac{0.586 \ ext{g}}{4.00 \ ext{g/mol}} \approx 0.1465 \ ext{mol} \)
This simple operation allows us to use the Ideal Gas Law later to find other properties of the gas such as volume.
Role of Vapor Pressure
Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid. At a specific temperature, a liquid like water will have a certain vapor pressure. This plays a key role in gas collection methods where gases are collected over water.
In this example, the water vapor has a pressure of 23.8 torr at 25掳C, which is equivalent to 0.03132 atm when converted into atmospheres. This pressure is part of the total pressure measured.
Understanding vapor pressure is essential because it affects the calculation of a gas's partial pressure in the mixture, directly influencing the results obtained using the Ideal Gas Law.
Temperature Conversion
Temperature is a crucial factor in gas law calculations, and converting Celsius to Kelvin is essential because the Ideal Gas Law uses Kelvin.
The Kelvin scale starts at absolute zero, which allows us to avoid negative temperatures in calculations. To convert from Celsius to Kelvin, simply add 273.15:
  • Temperature in Kelvin = Temperature in 掳C + 273.15
  • Example: \( 25^{\circ} \text{C} = 25 + 273.15 = 298.15 \text{K} \)
Just remember: Kelvin is king when it comes to using the Ideal Gas Law! This conversion is necessary for accurate mathematical modeling in thermodynamics and other physical science applications.

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Most popular questions from this chapter

For scuba dives below 150 \(\mathrm{ft}\) , helium is often used to replace nitrogen in the scuba tank. If 15.2 \(\mathrm{g}\) of \(\mathrm{He}(g)\) and 30.6 \(\mathrm{g}\) of \(\mathrm{O}_{2}(g)\) are added to a previously evacuated 5.00 \(\mathrm{L}\) tank at \(22^{\circ} \mathrm{C},\) calculate the partial pressure of each gas present as well as the total pressure in the tank.

Consider the unbalanced chemical equation below: $$\mathrm{CaSiO}_{3}(s)+\mathrm{HF}(g) \longrightarrow \mathrm{CaF}_{2}(a q)+\mathrm{SiF}_{4}(g)+\mathrm{H}_{2} \mathrm{O}( )$$ Suppose a 32.9 -g sample of \(\mathrm{CaSiO}_{3}\) is reacted with 31.8 \(\mathrm{L}\) of \(\mathrm{HF}\) at \(27.0^{\circ} \mathrm{C}\) and 1.00 \(\mathrm{atm}\) . Assuming the reaction goes to completion, calculate the mass of the \(\mathrm{SiF}_{4}\) and \(\mathrm{H}_{2} \mathrm{O}\) produced in the reaction.

Which noble gas has the smallest density at STP? Explain

A compressed gas cylinder contains \(1.00 \times 10^{3} \mathrm{g}\) argon gas. The pressure inside the cylinder is \(2050 .\) psi (pounds per square inch) at a temperature of \(18^{\circ} \mathrm{C}\) . How much gas remains in the cylinder if the pressure is decreased to 650 . psi at a temperature of \(26^{\circ} \mathrm{C} ?\)

A 20.0 -L stainless steel container at \(25^{\circ} \mathrm{C}\) was charged with 2.00 atm of hydrogen gas and 3.00 atm of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at \(25^{\circ} \mathrm{C}\) ? If the exact same experiment were performed, but the temperature was \(125^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C},\) what would be the pressure in the tank?
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