2.1: PASS Ideal Gases- General gas law calculation, changing temperature (2.E.12)

Exercise 2.E.12

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

Answer

Final pressure = 4.47 atm

Strategy Map

Step	Hint
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.	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.	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

\(\frac{P_i V_i}{n_i T_i}=\frac{P_f V_f}{n_f T_f}\)

\(\frac{P_i T_f}{T_i}=P_f \\\)

\(\begin{aligned}
& \text { Torr } \rightarrow \text { atmospheres } \\
& \qquad \begin{array}{l}
\frac{1344 \text { torr }}{1} \times \frac{1 \text { atm }}{760 \text { torr }}=\mathbf{1 . 7 6 8}
\end{array} \\
& \text { Degrees Celsius } \rightarrow \text { Kelvin } \\
& \qquad \begin{array}{l}
23^{\circ} \mathrm{C}+273.15=\mathbf{2 9 6} \boldsymbol{k} \\
475^{\circ} \mathrm{C}+273.15=\mathbf{7 4 8} \boldsymbol{k}
\end{array}
\end{aligned}\)

\(\begin{gathered}
\frac{P_i T_f}{T_i}=P_f \\
\frac{(1.77 \mathrm{~atm})(748 \mathrm{k})}{296 \mathrm{k}}=P_f \\
\boldsymbol{P}_{\boldsymbol{f}}=4.47 \mathrm{~atm}
\end{gathered}\)

Guided Solution

Download Guided Solution as a pdf

Guided Solution	Hint
This is a calculation type problem where we use the ideal gas laws to calculate how the pressure would change due to a change in conditions.	Refer to: LibreText Section 2.3
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?	We are told that the initial pressure is 1344 torr and that the initial temperature is 23°C. We are also given the final temperature which is 475°C. We also know that the amount of gas inside the can stays constant before and after we are evaluating its pressure.
The question asks you to find the pressure after the can is heated.
If the can is being heated you can expect the final pressure to be higher than the initial pressure.	If your calculated final pressure is lower than the initial pressure, you must have made an error somewhere.
Recall the ideal gas equation. \(P V=n R T\) How can we manipulate it to find the information we want?	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. \(\frac{P_i V_i}{n_i T_i}=R=\frac{P_f V_f}{n_f T_f}\) \(\frac{P_i V_i}{n_i T_i}=\frac{P_f V_f}{n_f T_f}\) Now we can get rid of values that stay constant during this problem. In this case the Volume and number of moles stays constant so it can be removed. (it can be divided out since it is the same on both sides). \(\frac{P_i T_f}{T_i}=P_f \\\) 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 are using values with the correct units. If necessary, use conversion factors.	Since we want our final pressure in atmospheres we will convert the initial pressure to atmospheres. Since the pressure of an ideal gas is proportional to temperature in Kelvin, we must convert our temperatures to Kelvin in order to use our derived relationship.
Complete Solution Conversions: \(\begin{aligned} & \text { Torr } \rightarrow \text { atmospheres } \\ & \qquad \begin{array}{l} \frac{1344 \text { torr }}{1} \times \frac{1 \text { atm }}{760 \text { torr }}=\mathbf{1 . 7 6 8} \end{array} \\ & \text { Degrees Celsius } \rightarrow \text { Kelvin } \\ & \qquad \begin{array}{l} 23^{\circ} \mathrm{C}+273.15=\mathbf{2 9 6} \boldsymbol{k} \\ 475^{\circ} \mathrm{C}+273.15=\mathbf{7 4 8} \boldsymbol{k} \end{array} \end{aligned}\) Manipulating the Ideal gas equation: \(P V=n R T\) R is a constant, we can rearrange and isolate ‘R.’ Since it is a constant, the initial conditions and final conditions will both equal the ‘R.’ \(\frac{P_i V_i}{n_i T_i}=R=\frac{P_f V_f}{n_f T_f}\) Since they both equal ‘R’ they also equal each other. \(\frac{P_i V_i}{n_i T_i}=\frac{P_f V_f}{n_f T_f}\) 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 simply remove them from our equation. \(\frac{P_i}{T_i}=\frac{P_f}{T_f}\) Now all that’s left is to isolate the desired variable, Pf. \(\frac{P_i T_f}{T_i}=P_f \\\) Finally, plug your initial and final temperatures and initial pressure into the derived equation, and solve for teh final pressure: \(\begin{gathered} \frac{P_i T_f}{T_i}=P_f \\ \frac{(1.77 \mathrm{~atm})(748 \mathrm{k})}{296 \mathrm{k}}=P_f \\ \boldsymbol{P}_{\boldsymbol{f}}=4.47 \mathrm{~atm} \end{gathered}\)
Check your work! We expected the final pressure to be higher than the initial pressure, and it is: Pi = 1.77 atm, Pf = 4.47 atm Why does this answer make chemical sense? We know from Amonton’s Law that temperature and pressure are proportional, meaning as temperature increases, the pressure will also increase. Using this knowledge, we know that our calculation must show that the final pressure is higher than the initial pressure.	If our calculation shows the final pressure as unchanged or less than the initial pressure, an error likely occurred in the calculation process.

(question source from page titled 9.E:Gases (Exercises) https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_(OpenSTAX)/09%3A_Gases/9.E%3A_Gases_(Exercises), shared under a CC BY 4.0 license, authored, remixed, and/or curated by OpenStax, original source https://openstax.org/books/chemistry/pages/9-exercises, Access for free at https://openstax.org/books/chemistry/pages/1-introduction)