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).

(T₂+273) K=200 kPa / 500 kPa ×1×(400+273) K

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).

(T₂+273) K=200 kPa / 500 kPa ×1×(400+273) K

Accepted Answer

Hey everyone. Let's go through this practice problem. Create a problem statement for the situation represented by the equation. And then I'm just going to rewrite it for readability. T sub two plus 273. Kelvins is equal to 300 elo pascals divided by 700 helo pascals multiplied by one multiplied by 350 plus 273 Kelvins. And we're given four options to choose from option. A calculate the final temperature attained by a gas. If it is expanded from 300 kg pascals to 700 kg pascals keeping the density constant, the initial temperature of the gas is 350 degrees Celsius. Option C calculate the final temperature attained by a gas if it is compressed from 700 kg pascals to 300 kg pascals keeping its density constant. And the same initial temperature option C calculate the final temperature if it is compressed from 700 kilobytes scales to 300 kilowatt scales keeping the volume constant. Same initial temperature. And option D calculate the final temperature attained by a gas. If it is expanded from 300 kg pascals to 700 kg pascals keeping its volume constant and the same initial temperature of 350 degrees Celsius. So this is kind of an unusual problem. But essentially what's happening here is that we have to pick the statement that best describes the equation that we were given. And all the statements relate to the relationship between an initial pressure and a final pressure or uh the change in temperatures and whatnot and, and basically variables that are related to the gas variables and specifically the ones that appear in various versions of the ideal gas law. So if we want to relate those variables between two different states of a gas, one thing that we can use is the I is the combined version of the ideal gas law which states that P sub one, the initial pressure multiplied by the initial volume divided by the initial temperature is equal to the final temperature. The final pressure P sub two multiplied by the final volume V sub two divided by the final temperature. And the equation that the problem gives us, let me fix this t the equation that the problem gives us features T sub two, the final temperature on one side of the equation. So if we want to see how the variables that are mentioned in the problem statement options and how those statement, how those numbers compare to the values in the equation. One thing we can do is solve the ideal gas law for T sub two. So I'm just going to algebraically solve the combined gas law for T sub two. And we can do this by multiplying both sides of the equation by T sub two. And then dividing both sides of the equation by P sub one V sub one divided by T sub one. And what we find is that T sub two is equal to the pressure. Sub two, P sub two divided by P sub one multiplied by V sub two, divided by V sub one multiplied by T sub one. And now if we compare the form of our new T sub two equation to the form of the T sub two equation given by the problem. Now it's easier to see what the variables correspond to. So we can see in the problem in the statement that was given to us, we have a 300 kg pascals divided by 700 kg pascals. And if we look at our new equation for T sub two, we have a P sub two divided by P sub one. So what that must mean is that P sub two must correspond to the one in the numerator. So P sub two is equal to 300 kg pascals and P sub one corresponds to the one in the denominator. So P sub one is 700 kg pascals. That means that the gas initial pressure was 700 kg pascals and then it's getting compressed to a final pressure of 300 kg pascals. We can also see that the temperature on the right side of the equation given 350 plus 2 73 Kelvins would correspond to T sub one, the initial temperature. So T sub one is equal to 350 degrees Celsius. And one thing that might be more interesting is the fact that the equation given the problem doesn't include any terms that look like a volume term. Though the problem does mention that there is a multiplied by one in there somewhere. So from that, the implication seems to be that we can, we can infer that the initial volume is equal to the final volume. Since the only way for the volume to not show up is if the same volume B sub two was being divided by an equal volume V sub one. So the volume must be constant during the process described in the problem. So we're going to have to pick a problem statement that involves keeping the volume constant. And so if we consider everything we know, there's only one option that seems to fit, we know that the initial pressure is 700 kg pascals and then it gets compressed to 300. And we also know that the volume is going to be kept constant. And the only option that has all of this information is option C which says calculate the final temperature attained by the gas if it was compressed from kg pascals to 300 kg pascals keeping the volume constant with an initial temperature of 350 degrees Celsius. So option C is the answer to this problem and that is all for this video. I hope this video helped you out. If you think you need more practice, please check out some of our other tutoring 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). (T₂+273) K=200 kPa / 500 kPa ×1×(400+273) K

Verified Solution

Key Concepts

Ideal Gas Law

Pressure Units

Temperature Conversion