In Problems 67, 68, 69, and 70 you are given the equation(s) used...

Question

In Problems 67, 68, 69, and 70 you are given the equation(s) used to solve a problem. For each of these, you are to

a. Write a realistic problem for which this is the correct equation(s).

p₂=300 cm^3/ 100 cm^3 ×1×2 atm

Accepted Answer

Hey everyone. Let's go through this practice problem. Create a question depicting a practical scenario for the following equation. And then I'm just going to rewrite it in a somewhat more readable way. In this P sub two is equal to cubic centimeters divided by 150 cubic centimeters, multiplied by one, multiplied by 3.5 atmospheres and were given four multiple choice options to choose from each one describing a slightly different scenario for the change of a gas option. A the volume of a gas decreases from 500 cubic centimeters to 100 50 cubic centimeters as it undergoes isothermal compression, determine the final pressure of the gas. If its initial pressure is 3.5 atmospheres. The statement in option B describes everything is the same except it's an isobaric compression instead of isothermal. The problem statement in option C describes the volume as an increase from 100 and 50 cubic centimeters to 500 cubic centimeters under isothermal expansion. And statement D is the same as statement C except it's an isobaric expansion instead of an isothermal one. This is kind of an unusual problem. But the basic idea of it is that we just want to pick this statement that best describes the equation that we were given that equation being one that relates the pressure with things like uh volume. With all the statements we have to choose from also including uh mentions of temperature uh with isothermal compressions or isobaric compressions. So we're gonna want to find some relationship between the different pressure and volume variables in order for us to see how the vari how the numbers given in the equation we were given line up with that. And the easiest way we can do that is by using the combined form of the ideal gas law, which states that the initial pressure P sub one multiplied by the initial volume divided by the initial temperature is equal to the final pressure P sub two multiplied by the final volume divided by the final temperature. Now, we can see from the equation given to us in the problem statement that the equation shown is an equation for P sub two. So if you want to get an equation that will match the form of the one we were given, let's take our combined ideal gas law formula and solve it for P sub two. So this is just a pretty simple algebraic computation we can solve for P sub two by multiplying both sides of the equation by T sub two and then dividing both sides of the equation by V sub two. And what we find is that P sub two is equal to V sub one divided by V sub two multiplied by T sub two, divided by T sub one multiplied by the initial pressure P sub one. Now right off the bat, if we compare the form to our new P sub two equation compare that form to what we were given in the problem statement. It's easy to see now how some of the variables match up with the numbers that we were given. So because we have a V sub one divided by V sub two in our new equation. And the problem statement has a 500 cubic centimeters divided by 100 and 50 cubic centimeters. That implies that V sub one, since it's in the numerator must be equal to the volume in the numerator of the given equation. So V sub one is 500 cubic centimeters and the final volume V sub two is equal to the one in the denominator 100 50 cubic centimeters. We can also see that since there is a P sub one on the right hand side of the right side of the pressure two equation that lines up with the 3. atmospheres that we were given in the problem statement. So P sub one is equal to 3.5 atmospheres. Now, the equation we found for P sub two also includes a final temperature T sub two divided by initial temperature T sub one term. But if we look at the problem statements equation, we don't have a term like that. Nothing here has units of, of Kelvins for temperature. But we are given a multiplied by one term in there, which to me seems pretty clearly like a clue that the reason why there's no temperature is because the temperatures are going to cancel out. And as we can see from the form of the equation we found earlier, the only way that can happen is if T sub two is equal to T sub one. In other words, the temperatures stay the same. In both cases, the temperature doesn't change. And recall that when we have a process where the temperature is constant, that means that the process is considered isothermal because the temperature stays the same. And now the information that we found is gonna be enough to narrow down the correct answer. So if we look at our options, we can see that. Well, first off, we know from what we've discovered that it's going to be an isothermal process. So we can cross out options B and C and, or B and D because they describe the process as being isobaric rather than isothermal. And we also found that the initial volume is going to be 500 cubic centimeters and then it compresses down to 100 50 cubic centimeters. And so the only option that really fits here is option A since that's the choice that describes the volume decreasing from 500 to 100 50 cubic centimeters with an isothermal compression and the final pressure is 3.5 atmospheres as we also discussed. And so that is the answer to this problem. So I hope this video helped you out. If you think you need more practice, please check out some of our other videos which will give you more experience with these types of problems. But that's all for now. I hope you all have a lovely day. Bye bye.

In Problems 67, 68, 69, and 70 you are given the equation(s) used to solve a problem. For each of these, you are to a. Write a realistic problem for which this is the correct equation(s). p₂=300 cm^3/ 100 cm^3 ×1×2 atm

Key Concepts

Gas Laws

Pressure and Volume Relationship

Units of Measurement

Watch next