Our Discord hit 10K members! 🎉 Meet students and ask top educators your questions.Join Here!

An aerosol can contains 400. mL of compressed gas at 5.20 atm. When all of the gas is sprayed into a large plastic bag, the bag inflates to a volume of 2.14 L. What is the pressure of gas in the plastic bag? Assume a constant temperature.

$$\begin{array}{l}{\text { Here, the temperature is constant and the amount of aerosol used is constant i.e. number of }} \\ {\text { moles of aerosol is constant. }} \\ {\text { As per Boyle's law, at given temperature and number of moles, pressure of the gas is inversely }} \\ {\text { proportional to volume. }} \\ {\text { P } \alpha \frac{1}{V}} \\ {P=\frac{\text { Constant }}{V}} \\ {\text { PV }=\text { Constant }} \\ {P_{1} V_{1}=P_{2} V_{2}=\text { Constant }}\end{array}$$$$\begin{array}{l}{\text { Where, } P_{1}, V_{1} \text { represents initial pressure and volume and } P_{2}, V_{2} \text { represent final pressure and }} \\ {\text { volume. }} \\ {\text { Given that an aerosol can contains } 400 \mathrm{mL}\left(\mathrm{V}_{1}\right) \text { of compressed gas at } 5.20 \mathrm{atm}\left(\mathrm{P}_{1}\right) \text { . It is }} \\ {\text { sprayed into a plastic bag. The bag inflates to a volume of } 2.14 \mathrm{L}\left(\mathrm{V}_{2}\right) . \text { Substitute these values in }} \\ {\text { the above equation. }} \\ {\qquad 2.14 \mathrm{L} \times 5.20 \mathrm{atm}} \\ {=0.97196 \mathrm{atm}} \\ {=0.972 \mathrm{atm}}\end{array}$$$$\text { The final pressure of the gas in the plastic bag is about } 0.972 \text { atm }$$

Kinetics

Gases

Discussion

You must be signed in to discuss.
TD

Carleton College

Stephen P.

Drexel University

Allea C.

University of Maryland - University College

KS

Video Transcript

problem. 47 says that an aerosol can containing 400 milliliters of compressed gas at 5.2 atmospheres is expelled into a large plastic bag with a volume of 2.14 leaders. So this problem asks you what the pressure of the gas is once it leaves, you can into the plastic bag. So the way that you can solve this iss, Um uh, by using properties of gas where, um, the the pressure and the volume, um, scale together and are equivalent across two systems if you're, you know, talking about the same the same gas. So in this case, it's almost exactly like I'm familiar with the c one of you once he to be thio equation. Where if you're making certain chemicals in the lab, um, the ultimately the concentration in the volume from one solution to the other, um have, uh, proportional relationships. So we can use this equation then and the information given to us from the problem thio to solve essentially for, um, uh, P, too. So in the aerosol can we know that the pressure is 5.20 atmospheres and we know that the can has a 400 militar volume. So the first thing that we can do here is convert this Miller leaders two liters by introducing this conversion factor, you know, that we want to solve for, um, Pete to here. And we have a volume of this plastic bag at 2.14 liters. So basically, uh, we can just solve through, um by dividing each side by 2.14 leaders giving us at the pressure of this plastic bag or pee too. ISS 0.972 atmospheres.

University of Pittsburgh - Main Campus

Topics

Kinetics

Gases

TD

Carleton College

Stephen P.

Drexel University

Allea C.

University of Maryland - University College

KS