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Chapter 10: Problem 40

An aerosol spray can with a volume of 250 \(\mathrm{mL}\) contains 2.30 \(\mathrm{g}\) of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) as a propellant. (a) If the can is at \(23^{\circ} \mathrm{C}\) , what is the pressure in the can? (b) What volume would the propane occupy at STP? (c) The can's label says that exposure to temperatures above \(130^{\circ}\) F may cause the can to burst. What is the pressure in the can at this temperature?

Short Answer

Expert verified
(a) The pressure in the can at 23掳C is 5.09 atm. (b) At STP, the propane gas occupies a volume of 1.18 L. (c) At 130掳F, the pressure in the can is 7.07 atm.

Step by step solution

01

(a) Pressure in the can at 23掳C

First, we need to convert the gas mass (2.30 g) to moles using propane's molar mass: Molar mass of propane (C3H8) = 3(12.01 g/mol) + 8(1.01 g/mol) = 44.11 g/mol Number of moles (n) = (2.30 g) / (44.11 g/mol) = 0.0521 mol Next, we need to convert the given temperature (23掳C) to Kelvin: T = 23掳C + 273.15 = 296.15 K Now, let's find the pressure using the Ideal Gas Law. We need to convert the volume to Liters first: V = 250 mL / 1000 = 0.250 L The ideal gas constant (R) in L路atm/mol路K = 0.0821 L路atm/mol路K Now, we plug the values into the Ideal Gas Law to find the pressure (P): PV = nRT P = nRT / V P = (0.0521 mol)(0.0821 L路atm/mol路K)(296.15 K) / (0.250 L) P = 5.09 atm The pressure in the can is 5.09 atm at 23掳C.
02

(b) Volume of propane gas at STP

Standard Temperature and Pressure (STP) conditions are 0掳C (273.15 K) and 1 atm. Using the Ideal Gas Law, we can calculate the volume at STP: \(V_{STP} = \frac{nRT_{STP}}{P_{STP}}\) \(V_{STP} = \frac{(0.0521 \text{ mol})(0.0821 \text{ L路atm/mol路K})(273.15 \text{ K})}{1 \text{ atm}}\) \(V_{STP} = 1.18 \text{ L}\) At STP, the propane gas occupies a volume of 1.18 L.
03

(c) Pressure in the can at 130掳F

First, convert the given temperature (130掳F) to Kelvin: \(T_{new} = \frac{5}{9}(130 - 32) + 273.15 = 324.15 K\) Now, apply the Ideal Gas Law to find the pressure in the can at this new temperature: \(P_{new} = \frac{nRT_{new}}{V}\) \(P_{new} = \frac{(0.0521 \text{ mol})(0.0821 \text{ L路atm/mol路K})(324.15 \text{ K})}{0.250 \text{ L}}\) \(P_{new} = 7.07 \text{ atm}\) At a temperature of 130掳F, the pressure in the can is 7.07 atm.

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Key Concepts

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

Pressure Calculations
In chemistry, understanding how to calculate pressure inside a container is pivotal, especially when dealing with gases in enclosed spaces like aerosol cans. The Ideal Gas Law is a fundamental formula used for pressure calculations. It combines variables such as the number of moles ( ext{n}), the gas constant ( ext{R}), volume ( ext{V}), and temperature ( ext{T}) to find the pressure ( ext{P}). This relationship is expressed as:
\[P = \frac{nRT}{V}\]
To practically apply this formula, you need:
  • Converting the mass of the gas to moles by using the molar mass of the substance.
  • Adjusting the temperature into Kelvin since the Ideal Gas Law requires it in this unit.
  • Converting volume into liters when dealing with measurements given in milliliters or cubic centimeters.
Understanding how to apply these conversions is essential when solving problems involving pressure in gases. For instance, knowing the pressure inside an aerosol can at a specific temperature helps determine safety precautions鈥攆or example, the risk of bursting at high temperatures.
Molar Mass
Molar mass is a critical concept when working with gases, as it helps convert the gas's mass into moles. The molar mass is essentially the mass of one mole of a substance and is usually expressed in grams per mole (\text{g/mol}). For propane (\text{C}_3\text{H}_8), you calculate it by adding up the atomic masses of three carbon atoms and eight hydrogen atoms leading to:
  • Carbon: 3 atoms 脳 12.01 \text{g/mol} = 36.03 \text{g/mol}.
  • Hydrogen: 8 atoms 脳 1.01 \text{g/mol} = 8.08 \text{g/mol}.
  • Total Molar Mass: 44.11 \text{g/mol}.

To convert the mass of propane into moles, you divide the mass of the gas by its molar mass:\[n = \frac{\text{mass of gas}}{\text{molar mass}}\]Using the correct molar mass in calculations is crucial to ensure the accuracy of results in chemical equations and reactions.
STP Conditions
Standard Temperature and Pressure (STP) conditions are a set of predefined conditions used as a reference point in chemistry. At STP, a gas is at a temperature of 0掳C (273.15 K) and a pressure of 1 atm. These conditions are used to provide a standard comparison for gas volumes.
At STP, 1 mole of an ideal gas occupies a volume of 22.4 liters. This uniformity simplifies comparing gases' behavior under different conditions.
When calculating the volume of a gas at STP using the Ideal Gas Law, it's essential to adjust the known values to these specific conditions:
\[V_{STP} = \frac{nRT_{STP}}{P_{STP}}\]
Understanding STP conditions helps solve problems involving gases by providing a reliable baseline of measurement, especially when converting conditions from other environments back to standard ones. It also aids in understanding how gases expand or contract with changes in temperature and pressure.

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Most popular questions from this chapter

Which of the following statements is false? \begin{equation}\begin{array}{l}{\text { (a) Gases are far less dense than liquids. }} \\ {\text { (b) Gases are far more compressible than liquids. }} \\\ {\text { (c) Because liquid water and liquid carbon tetrachloride do }} \\\ {\text { not mix, neither do their vapors. }} \\ {\text { (d) The volume occupied by a gas is determined by the volume }} \\ {\text { of its container. }}\end{array}\end{equation}

A quantity of \(\mathrm{N}_{2}\) gas originally held at 5.25 atm pressure in a 1.00 -L container at \(26^{\circ} \mathrm{C}\) is transferred to a \(12.5-\mathrm{L}\) container at \(20^{\circ} \mathrm{C}\) . A quantity of \(\mathrm{O}_{2}\) gas originally at 5.25 atm and \(26^{\circ} \mathrm{C}\) in a \(5.00-\mathrm{L}\) container is transferred to this same container. What is the total pressure in the new container?

Table 10.3 shows that the van der Wals \(b\) parameter has units of L/mol. This means that we can calculate the sizes of atoms or molecules from the \(b\) parameter. Refer back to the discussion in Section \(7.3 .\) Is the van der Waals radius we calculate from the \(b\) parameter of Table 10.3 more closely associated with the bonding or nonbonding atomic radius discussed there? Explain.

Chlorine is widely used to purify municipal water supplies and to treat swimming pool waters. Suppose that the volume of a particular sample of \(\mathrm{Cl}_{2}\) gas is 8.70 \(\mathrm{L}\) at 895 torr and \(24^{\circ} \mathrm{C}\) .(a) How many grams of \(\mathrm{Cl}_{2}\) are in the sample? (b) What volume will the \(\mathrm{Cl}_{2}\) occupy at \(\mathrm{STP}\) ? (c) At what temperature will the volume be 15.00 \(\mathrm{L}\) if the pressure is \(8.76 \times 10^{2}\) torr? (d) At what pressure will the volume equal 5.00 L if the temperature is \(58^{\circ} \mathrm{C}\) ?

Suppose you have two 1 -L flasks, one containing \(\mathrm{N}_{2}\) at STP, the other containing \(\mathrm{CH}_{4}\) at STP. How do these systems compare with respect to (a) number of molecules, (b) density, (c) average kinetic energy of the molecules, (d) rate of effusion through a pinhole leak?
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