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Chapter 10: Problem 40

An aerosol spray can with a volume of \(125 \mathrm{~mL}\) contains \(1.30 \mathrm{~g}\) of propane gas \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) as a propellant. (a) If the can is at \(25^{\circ} \mathrm{C}\), what is the pressure in the can? (b) What volume would the butane occupy at \(S T P ?\) (c) The can's label says that exposure to temperatures above \(50^{\circ} \mathrm{C}\) may cause the can to burst. What is the pressure in the can at this temperature?

Short Answer

Expert verified
The pressure in the can at 25掳C is found using the Ideal Gas Law equation rearranged for P and plugging in the calculated values of n, V, and T: \[P = \frac{(1.3\mathrm{~g}/44.097\mathrm{~g/mol})\times(8.314\mathrm{~J/[mol\cdot K]})\times(298.15\mathrm{~K})}{125\mathrm{~mL}}\] The volume of propane at STP is calculated using the Ideal Gas Law equation rearranged for V and plugging in the values for n, R, and T: \[V_{STP} = \frac{(1.3\mathrm{~g}/44.097\mathrm{~g/mol})\times(8.314\mathrm{~J/[mol\cdot K]}/101.325\mathrm{~J/[L\cdot atm]})\times(273.15\mathrm{~K})}{1\mathrm{~atm}}\] Lastly, the pressure in the can at 50掳C is calculated using the Ideal Gas Law equation rearranged for P and plugging in the calculated values of n, V, and T': \[P' = \frac{(1.3\mathrm{~g}/44.097\mathrm{~g/mol})\times(8.314\mathrm{~J/[mol\cdot K]})\times(323.15\mathrm{~K})}{125\mathrm{~mL}}\]

Step by step solution

01

Calculate the number of moles of propane gas

First, we need to determine the number of moles of propane gas in the aerosol can. To do this, we will use the molar mass of propane gas (C3H8) which is approximately \(44.097 \mathrm{~g/mol}\). To find the number of moles, we divide the mass of the propane gas by its molar mass: \[n = \frac{1.3\mathrm{~g}}{44.097\mathrm{~g/mol}}\]
02

Determine the temperature in Kelvin

The Ideal Gas Law equation requires temperatures to be expressed in Kelvin. Therefore, we need to convert the given temperatures (25掳C and 50掳C) to Kelvin. To convert Celsius temperature to Kelvin, we use the following formula: \[T(K) = T(掳C) + 273.15\] For 25掳C: \[T = 25 + 273.15 = 298.15\mathrm{~K}\] And for 50掳C: \[T' = 50 + 273.15 = 323.15\mathrm{~K}\]
03

Calculate the pressure in the can at 25掳C

Now that we have the number of moles and temperature in Kelvin, we can plug these values into the Ideal Gas Law equation alongside the can's volume to find the pressure at 25掳C. \(PV = nRT\) We will rearrange the equation for P: \[P = \frac{nRT}{V}\] Using the values for n, V, and T: \[P = \frac{(1.3\mathrm{~g}/44.097\mathrm{~g/mol})\times(8.314\mathrm{~J/[mol\cdot K]})\times(298.15\mathrm{~K})}{125\mathrm{~mL}}\]
04

Calculate the volume of propane at STP

Now, we will calculate the volume that the propane would occupy at STP (Standard Temperature and Pressure). The standard conditions for temperature and pressure are T = 273.15 K and P = 1 atm. Using the Ideal Gas Law, we will solve for the volume at STP: \[V_{STP} = \frac{nR_{STP}T_{STP}}{P_{STP}}\] Where \(R_{STP} = \frac{R}{101.325\mathrm{~J/[L\cdot atm]}}\) Now we plug in the values for n, R, and T: \[V_{STP} = \frac{(1.3\mathrm{~g}/44.097\mathrm{~g/mol})\times(8.314\mathrm{~J/[mol\cdot K]}/101.325\mathrm{~J/[L\cdot atm]})\times(273.15\mathrm{~K})}{1\mathrm{~atm}}\]
05

Calculate the pressure in the can at 50掳C

Lastly, we will find the pressure in the aerosol can at an increased temperature of 50掳C (323.15 K) using the Ideal Gas Law equation: \[P' = \frac{nRT'}{V}\] Using the calculated values of n, V, and T': \[P' = \frac{(1.3\mathrm{~g}/44.097\mathrm{~g/mol})\times(8.314\mathrm{~J/[mol\cdot K]})\times(323.15\mathrm{~K})}{125\mathrm{~mL}}\] Now the student can plug in the given values for each step and solve for the pressure at 25掳C, the volume at STP, and the pressure at 50掳C.

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Key Concepts

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

Moles of Gas
To calculate the number of moles of propane gas, you need to know the mass of the gas and its molar mass. Moles give us a way to express amounts of a substance using the number of particles. For propane \((C_3H_8)\), the molar mass is approximately \(44.097 \, \text{g/mol}\).

Here's the formula to find the number of moles:
  • Divide the mass of the propane by its molar mass.
Given:
Mass of propane \(= 1.30 \, \text{g}\)
Molar mass \(= 44.097 \, \text{g/mol}\)
The number of moles \(n\) is calculated as: \[n = \frac{1.3}{44.097} \, \text{mol}\] This calculation tells you how much of the gas is present in terms of moles.
Conversion to Kelvin
When dealing with gas laws, temperature must be in Kelvin because it avoids negative numbers and reflects an absolute scale. Converting Celsius to Kelvin is straightforward.

Here's how you convert:
  • Add 273.15 to the Celsius temperature.
For \(25^{\circ} \text{C}:\)
\[T = 25 + 273.15 = 298.15 \, \text{K}\] For \(50^{\circ} \text{C}:\)
\[T' = 50 + 273.15 = 323.15 \, \text{K}\]
Making sure your temperatures are in Kelvin ensures accurate calculations when using the Ideal Gas Law.
Standard Temperature and Pressure
Standard Temperature and Pressure (STP) is used as a reference in scientific calculations. It's defined as a temperature of \(273.15 \, \text{K}\) and pressure of \(1 \, \text{atm}\).

Using STP allows for consistent measurements and comparisons. At STP, a mole of any ideal gas occupies \(22.414 \, \text{L}\). This condition helps in understanding how gases behave in typical conditions. In problems, you often use STP to predict the behavior of gases when the temperature and pressure are standardized.
Pressure Calculation
Pressure calculation in gases can be done using the Ideal Gas Law: \(PV = nRT\). Here, \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature in Kelvin.

The rearranged formula for pressure is: \[P = \frac{nRT}{V}\] To find pressure at given conditions:
  • Plug in values for \(n, R, T, \) and \(V\).
  • Use \((8.314 \, \text{J/mol路K})\) for \(R\).
  • Check units; volume should be in \(\text{liters}\).
For example, calculate the pressure at \(25^{\circ} \text{C}\):
\[P = \frac{\left(\frac{1.3}{44.097}\right) \times 8.314 \times 298.15}{0.125}\]
This method helps quickly find the pressure of a gas under specific conditions.

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Most popular questions from this chapter

(a) The compound 1 -iodododecane is a nonvolatile liquid with a density of \(1.20 \mathrm{~g} / \mathrm{mL}\). The density of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\). What do you predict for the height of a barometer column based on 1 -iodododecane, when the atmospheric pressure is 749 torr? \((\mathbf{b})\) What is the pressure, in atmospheres, on the body of a diver if he is \(21 \mathrm{ft}\) below the surface of the water when the atmospheric pressure is 742 torr?

You have a gas at \(25^{\circ} \mathrm{C}\) confined to a cylinder with a movable piston. Which of the following actions would double the gas pressure? (a) Lifting up on the piston to double the volume while keeping the temperature constant; (b) Heating the gas so that its temperature rises from \(25^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\), while keeping the volume constant; (c) Pushing down on the piston to halve the volume while keeping the temperature constant.

The planet Jupiter has a surface temperature of \(140 \mathrm{~K}\) and a mass 318 times that of Earth. Mercury (the planet) has a surface temperature between \(600 \mathrm{~K}\) and \(700 \mathrm{~K}\) and a mass 0.05 times that of Earth. On which planet is the atmosphere more likely to obey the ideal-gas law? Explain.

Which statement concerning the van der Waals constants \(a\) and \(b\) is true? (a) The magnitude of \(a\) relates to molecular volume, whereas \(b\) relates to attractions between molecules. (b) The magnitude of \(a\) relates to attractions between molecules, whereas \(b\) relates to molecular volume. (c) The magnitudes of \(a\) and \(b\) depend on pressure. (d) The magnitudes of \(a\) and \(b\) depend on temperature.

The temperature of a \(5.00-\mathrm{L}\) container of \(\mathrm{N}_{2}\) gas is increased from \(20^{\circ} \mathrm{C}\) to \(250^{\circ} \mathrm{C}\). If the volume is held constant, predict qualitatively how this change affects the following: (a) the average kinetic energy of the molecules; (b) the rootmean- square speed of the molecules; (c) the strength of the impact of an average molecule with the container walls; \(\mathbf{d}\) ) the total number of collisions of molecules with walls per second.
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