Problem 40 - Chemistry: The Central Science

Question

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 says that exposure to temperatures above \(130^{\circ} \mathrm{F}\) may cause the can to burst. What is the pressure in the can at this temperature?

Step-by-Step Solution

Answer

(a) The pressure in the can at \(23^{\circ} \mathrm{C}\) is \(5.24\,\mathrm{atm}\). (b) The volume of propane at STP is \(0.233\,\mathrm{L}\). (c) The pressure in the can at \(130^{\circ} \mathrm{F}\) is \(5.87\,\mathrm{atm}\).

Step 1: Calculate the number of moles (n) of propane

Given the mass of propane as \(2.30\,\mathrm{g}\), we can find the number of moles by dividing the mass by the molar mass of propane. The molar mass of propane is \(3 \times (12.01\,\mathrm{g/mol}) + 8 \times (1.01\,\mathrm{g/mol}) = 44.11\,\mathrm{g/mol}\). Therefore, \( n = \frac{2.30\,\mathrm{g}}{44.11\,\mathrm{g/mol}} = 0.0521\,\mathrm{mol} \)

Step 2: Convert the given temperatures to Kelvin

To convert Celsius to Kelvin, add 273.15. For part (a), we're given a temperature of \(23^{\circ} \mathrm{C}\), so: \( T_1 = (23 + 273.15)\,\mathrm{K} = 296.15\,\mathrm{K} \) For part (c), we're given a temperature of \(130^{\circ} \mathrm{F}\). First, convert Fahrenheit to Celsius using the formula \(\frac{5}{9}(F - 32)\). Then, convert to Kelvin: \( T_2 = (130 - 32) \times \frac{5}{9} + 273.15 = 328.71\,\mathrm{K} \)

Step 3: Calculate the pressure inside the can at the given temperatures

We can now use the Ideal Gas Law (PV=nRT) to calculate the pressure in the can at the two given temperatures. (a) For \(T_1 = 296.15\,\mathrm{K}\), \( P_1 = \frac{nRT_1}{V} = \frac{0.0521\,\mathrm{mol} \times 0.0821\,\mathrm{L \cdot atm/mol \cdot K} \times 296.15\,\mathrm{K}}{0.250\,\mathrm{L}} = 5.24\,\mathrm{atm} \) (b) For \(T_2 = 328.71\,\mathrm{K}\), \( P_2 = \frac{nRT_2}{V} = \frac{0.0521\,\mathrm{mol} \times 0.0821\,\mathrm{L \cdot atm/mol \cdot K} \times 328.71\,\mathrm{K}}{0.250\,\mathrm{L}} = 5.87\,\mathrm{atm} \)

Step 4: Calculate the volume of propane at STP

Standard temperature and pressure (STP) is defined as \(0^{\circ} \mathrm{C}\) and 1 atmospheric pressure. We will use the combined gas law to find the volume at STP: \( \frac{P_1V_1}{T_1} = \frac{P_{STP}V_{STP}}{T_{STP}} \) To find the volume (V_{STP}): \(V_{STP} = \frac{P_1 V_1 T_{STP}}{P_{STP} T_1} = \frac{5.24\,\mathrm{atm} \times 0.250\,\mathrm{L} \times 273.15\,\mathrm{K}}{1\,\mathrm{atm} \times 296.15\,\mathrm{K}} = 0.233\,\mathrm{L}\) So: (a) The pressure in the can at \(23^{\circ} \mathrm{C}\) is \(5.24\,\mathrm{atm}\). (b) The volume of propane at STP is \(0.233\,\mathrm{L}\). (c) The pressure in the can at \(130^{\circ} \mathrm{F}\) is \(5.87\,\mathrm{atm}\).

Key Concepts

Gas Laws

The gas laws are a set of mathematical relationships that describe the behavior of gases. They are essential in understanding how gases interact under different conditions. One of the most important of these laws is the Ideal Gas Law, which is represented by the equation \( PV = nRT \).

In this equation:

• \(P\) stands for pressure
• \(V\) represents volume
• \(n\) is the number of moles of the gas
• \(R\) is the ideal gas constant \(0.0821 \, \text{L} \cdot \text{atm/mol} \cdot \text{K}\)
• \(T\) is the temperature in Kelvin

Understanding the Ideal Gas Law allows you to predict how changing one of these variables will affect another. For example, if you increase the temperature of a gas while keeping the number of moles and volume constant, the pressure will also increase. This ability to predict changes makes the Ideal Gas Law incredibly useful in various scientific and practical applications.

Molar Mass

Molar mass is crucial when dealing with chemical calculations, particularly in determining the amount of substances in reactions. It tells us the mass of a single mole of a substance. For instance, the molar mass of propane (\( \text{C}_3\text{H}_8 \)) is calculated by adding up the atomic masses of all the atoms in a molecule of propane.

Propane consists of three carbon atoms and eight hydrogen atoms. The atomic mass of carbon is \(12.01 \, \text{g/mol}\) and hydrogen is \(1.01 \, \text{g/mol}\). Therefore, the molar mass of propane is:

• \(3 \times 12.01 \, \text{g/mol} + 8 \times 1.01 \, \text{g/mol} = 44.11 \, \text{g/mol}\)

This means that one mole of propane weighs \(44.11 \, \text{g}\). Knowing the molar mass allows you to convert between the mass of a substance and the number of moles, which is essential for stoichiometric calculations in chemical reactions.

Standard Temperature and Pressure (STP)

Standard temperature and pressure (STP) is a reference point used in chemistry to provide a common ground for studying gases. It is defined as a temperature of \(0^{\circ} \, \text{C}\) (or \(273.15 \, \text{K}\)) and a pressure of \(1 \, \text{atm}\). Using STP allows scientists and engineers to compare results under consistent conditions.

At STP, one mole of an ideal gas occupies a volume of \(22.4 \, \text{L}\). This concept is useful for calculating gas volumes under different conditions through gas laws. For exercises involving STP, remember:

• Temperature = \(273.15 \, \text{K}\)
• Pressure = \(1 \, \text{atm}\)
• Volume of one mole of gas = \(22.4 \, \text{L}\)

This simplifies calculations by providing a standard baseline to compare how gases behave when deviating from STP conditions.

Conversion Between Units

Accurate scientific and chemical calculations often require converting between various units of measurement. A solid understanding of unit conversions ensures that computations are both accurate and meaningful. Two common temperature units in chemistry are Celsius (\(\degree\text{C}\)) and Kelvin (\(\text{K}\)).

To convert Celsius to Kelvin, simply add \(273.15\) to the Celsius temperature: \(T(\text{K}) = T(\degree\text{C}) + 273.15\).

For converting Fahrenheit to Celsius, use the formula:
\(T(\degree\text{C}) = \frac{5}{9}(T(\degree\text{F}) - 32)\)
Understanding these conversions is crucial for accurate data analysis and applying gas laws correctly, as they often require temperatures in Kelvin.

Chemical Calculations

Chemical calculations are the backbone of quantitative chemistry, involving the use of mathematical techniques to determine quantities of substances. They allow for the prediction and analysis of chemical reactions. When solving these calculations, the fundamental aspects often revolve around:

• Finding the amount of moles using the molar mass of substances
• Using the Ideal Gas Law to calculate pressures, volumes, and temperatures of gases
• Applying stoichiometry to deduce the reactants or products needed or produced

In this specific exercise, chemical calculations are employed to determine gas pressure changes using the Ideal Gas Law, as well as to convert conditions to STP to find the volume of a gas. Mastery in these calculations aids in solving real-world problems like gas storage, reactions, and how altering environmental conditions could impact them.