9 Ideal Gases — General Gas Law, Changing Temperatures

Question 

A spray can, which has a pressure of 1344 torr at 23°C, is used until it is empty, except for the propellent gas.

If the can is thrown into a fire at 475°C, what will be the pressure in the hot can in atmospheres?

 

Show/Hide Answer

Final pressure = 4.47 atm

Refer to Section 2.3 Relating Pressure, Volume, Amount, and Temperature – The Ideal Gas Law (1).

Strategy Map

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Check out the strategy map.

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Table 1: Strategy Map
Strategy Map Steps
1. Identify the information given in the question and what conditions are changing.
2. Identify what the question is asking you to calculate and what variable it would be represented by.
3. Choose an equation that compares the correct relationships.

Show/Hide Hint

You will need to manipulate the equation to compare the initial and final conditions.

4. Do any necessary conversions to ensure you are using the appropriate units.

Show/Hide Hint

Calculations for ideal gases using temperature must have the temperature in Kelvin.

Since we want our final pressure in atmospheres, we can convert the initial pressure to atmospheres in our calculation.

Solution

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Check out this solution.

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[latex]\begin{gathered} \frac{\mathrm{P}_\mathrm{i} \mathrm{V}_\mathrm{i}} {\mathrm{n}_\mathrm{i} \mathrm{T}_\mathrm{i}}=\frac{\mathrm{P}_\mathrm{f} \mathrm{V}_\mathrm{f}}{\mathrm{n}_\mathrm{f} \mathrm{T}_\mathrm{f}} \\ \\ \frac{\mathrm{P}_\mathrm{i} \mathrm{T}_\mathrm{f}}{\mathrm{T}_\mathrm{i}}=\mathrm{P}_\mathrm{f} \\ \\ \frac{1344 \mathrm{~torr}}{1} \times \frac{1 \mathrm{~atm}}{760 \mathrm {~torr }}=1 . 7 6 8\mathrm{~atm} \\ \\ 23^{\circ} \mathrm{C}+273.15=\mathrm{2 9 6 K} \\ \\ 475^{\circ} \mathrm{C}+273.15=\mathrm{7 4 8 K} \\ \\ \frac{\mathrm{P}_\mathrm{i} \mathrm{T}_\mathrm{f}}{\mathrm{T}_\mathrm{i}}=\mathrm{P}_\mathrm{f} \\ \\ \frac{(1.77 \mathrm{~atm})(748 \mathrm{k})}{296 \mathrm{k}}=\mathrm{P}_\mathrm{f}\\ \\ \mathrm{P}_{\mathrm{f}}=4 . 4 7 \mathrm {~atm} \end{gathered}[/latex]

Guided Solution

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The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.

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Table 2: Guided Solution
Guided Solution Ideas
This question is a calculation problem where we use the ideal gas laws to calculate how the pressure would change due to a change in conditions.

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Refer to Section 2.3 Relating Pressure, Volume, Amount, and Temperature – The Ideal Gas Law (1).

A spray can, which has a pressure of 1344 torr at 23°C, is used until it is empty except for the propellent gas. If the can is thrown into a fire at 475°C, what will be the pressure in the hot can in atmospheres?

The question asks you to find the pressure after the can is heated.

Show/Hide Think About This!

The question tells us that the initial pressure is 1344 torr and the initial temperature is 23°C. It also gives us the final temperature, which is 475°C.

We also know that the amount of gas inside the can stays constant before and after we evaluate its pressure.

If the can is heated, you can expect the final pressure to be higher than the initial pressure.

Show/Hide Watch Out!

Your calculated final pressure should be higher than the initial pressure. If it is not, then you must have made an error somewhere.

Recall the ideal gas equation.

Show/Hide Don’t Forget!

\begin{equation}
\mathrm{P V=n R T}
\end{equation}

How can we manipulate it to find the information we want?

Show/Hide Think About This!

We know that the ideal gas equation is PV = nRT, where ‘R’ is a constant. Because this value is constant, we can isolate it and create a derived equation where the initial values equal the final values.

\begin{gathered}
\frac{\mathrm{P}_\mathrm{i} \mathrm{V}_\mathrm{i}}{\mathrm{n}_\mathrm{i} \mathrm{T}_\mathrm{i}}=\mathrm{R}=\frac{\mathrm{P}_\mathrm{f} \mathrm{V}_\mathrm{f}}{\mathrm{n}_\mathrm{f} \mathrm{T}_\mathrm{f}} \\
\frac{\mathrm{P}_\mathrm{i} \mathrm{V}_\mathrm{i}}{\mathrm{n}_\mathrm{i} \mathrm{T}_\mathrm{i}}=\frac{\mathrm{P}_\mathrm{f} \mathrm{V}_\mathrm{f}}{\mathrm{n}_\mathrm{f} \mathrm{T}_\mathrm{f}}
\end{gathered}

Now, we can get rid of values that stay constant during this problem. In this case, the volume and number of moles remain constant; this means we can remove them. We can divide them out since they are the same on both sides.

\begin{equation}
\frac{\mathrm{P_i T_f}}{\mathrm{T_i}}=\mathrm{P_f}
\end{equation}

Once we isolate for our desired variable, we get our new equation, which we can plug in and use to solve for the final pressure.

Before plugging into your equation, ensure you use values with the correct units. If necessary, use conversion factors.

Show/Hide Don’t Forget!

Since we want our final pressure in atmospheres, we will convert the initial pressure to atmospheres.

The pressure of an ideal gas is proportional to the temperature in Kelvin. Because of this, we must convert our temperatures to Kelvin to use our derived relationship.

Table 3: Complete Solution
Complete Solution
Conversions:

Torr to atmospheres:

\begin{equation}
\frac{1344 \text { torr }}{1} \times \frac{1 \text { atm }}{760 \text { torr }}=1 . \mathrm{7 6}_8 \text { atm }
\end{equation}

Degree Celsius to Kelvin:

\begin{equation}
\begin{gathered}
23^{\circ} \mathrm{C}+273.15=\mathrm{2 9 6 K} \\
475^{\circ} \mathrm{C}+273.15=\mathrm{7 4 8 K}
\end{gathered}
\end{equation}

Manipulating the ideal gas equation:

\begin{equation}
\mathrm{P V=n R T}
\end{equation}

R is a constant, which means we can rearrange and isolate ‘R.’ Since it is a constant, the initial and final conditions will equal the ‘R.’

\begin{equation}
\frac{\mathrm{P_i V_i}}{\mathrm{n_i T_i}}=\mathrm{R}=\frac{\mathrm{P_f V_f}}{\mathrm{n_f T_f}}
\end{equation}

Since they both equal ‘R,’ they also equal each other.

\begin{equation}
\frac{\mathrm{P_i V_i}}{\mathrm{n_i T_i}}=\frac{\mathrm{P_f V_f}}{\mathrm{n_f T_f}}
\end{equation}

Now, we can cancel out the variables with values that do not change for this problem. In this case, the volume and number of moles do not change. We can remove them from our equation.

\begin{equation}
\frac{\mathrm{P_i}}{\mathrm{T_i}}=\frac{\mathrm{P_f}}{\mathrm{T_f}}
\end{equation}

Now, all we have to do is to isolate the desired variable, Pf.

\begin{equation}
\frac{\mathrm{P_i} \mathrm{T_f}}{\mathrm{T_i}}=\mathrm{P_f}
\end{equation}

Plugging into the derived equation:

\begin{equation}
\begin{gathered}
\frac{(1.77 \mathrm{~atm})(748 \mathrm{K})}{\mathrm{296 K}}=\mathrm{P_f} \\
\mathrm{P_f}=4.47 \mathrm{~atm}
\end{gathered}
\end{equation}

Check Your Work

If our calculation shows the final pressure as unchanged or less than the initial pressure, an error likely occurred in the calculation process.

Does your answer make chemical sense?

Show/Hide Answer

We expected the final pressure to be higher than the initial pressure, and it is:

Pi = 1.77 atm, Pf = 4.47 atm

Amonton’s Law says temperature and pressure are proportional, meaning that as temperature increases, pressure also increases. Using this knowledge, we know that our calculation must show that the final pressure is higher than the initial pressure.

PASS Attribution

References

1. OpenStax. 2.3: Relating Pressure, Volume, Amount, and Temperature – The Ideal Gas Law. In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520). LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/02%3A_Gases/2.03%3A_Relating_Pressure_Volume_Amount_and_Temperature_-_The_Ideal_Gas_Law.

2. Blackstock, L.; Brewer, S.; Jensen, A. In PASS Chemistry Book CHEM 1510/1520; LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510%2F%2F1520.

3. Blackstock, L.; Brewer, S.; Jensen, A. 2.1: PASS Ideal Gases- General gas law calculation, changing temperature (2.E.12). In PASS Chemistry Book CHEM 1510/1520. LibreTexts, 2024. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510%2F%2F1520/02%3A_Gases/2.01%3A_2.1_PASS_Ideal_Gases-_General_gas_law_calculation_changing_temperature_(2.E.12).

4. OpenStax. 9.E: Gases (Exercises). In Chemistry 1e (OpenSTAX). LibreTexts, 2023. https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX).

5. Flowers, P.; Robinson, W. R.; Langley, R.; Theopold, K. Ch. 9 Exercises. In Chemistry; OpenStax, 2015. https://openstax.org/books/chemistry/pages/9-exercises.

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