0.10 mol of argon gas is admitted to an evacuated 50 cm^3 container at 20°C. The gas then undergoes an isochoric heating to a temperature of 300°C. b. Show the process on a pV diagram. Include a proper scale on both axes.

Question

pV diagram showing the isochoric heating of argon gas in a 50 cm³ container.

Accepted Answer

Hey, everyone in this problem, we're told that a 125 mL drum holds 1250. moles of helium gas at 15 degrees Celsius. The drum allows iso chloric warming of the gass to 225 degrees Celsius. We asked to use a PV graph with correct values on the axes to represent the process. So instead of having multiple choice options, we're given the outline of a graph and we're asked to fill it in. So the first thing to note is that the problem tells us that the warming is iso choric. That means that the volume is going to be kept constant if the volume is constant. And that means our graph shouldn't show any volume variation at all. And since volume is what the horizontal axis is, that means that all of our points on the graph we draw should be at the same volume. This means that our graph should just look like a vertical line. The second thing to note is that the volume is actually given to us in the problem given to us as 125 mL. Now the graph shows the volume axis as being in cubic centimeters. But fortunately for us, one mL is equal to one cc. So that's a fairly easy conversion. But what that means is that if we follow the graph to find where the point is, where we have 100 25 cubic centimeters, then we can see that the point is something that already has a tick on it. And we know that the plot should just be a vertical line above this point. So our graph should look something like this. It should just look like a vertical line above the point where the volume is 125 cubic centimeters. However, we still need to find the initial and final pressures to represent the process. We can find a relationship between pressure and volume and temperature using the ideal gas law which states that the pressure multiplied by the volume of gas is equal to the number of moles multiplied by the ideal gas constant multiplied by its temperature. We can solve this for pressure by dividing both sides of the equation by volume. So the pressure of the gas is equal to the number of moles multiplied by the ideal gas constant multiplied by the temperature all divided by the gas as volume. The problem gives us both the initial and final temperatures. So let's use those to find the initial and final pressures. First, the initial temperature is given to us as 15 degrees Celsius, which if we add 273 to it. Then we can find this temperature in Kelvins as Kelvins. The number of moles N is given as 0.25 moles. And the volume was given to us as 100 25 cubic centimeters which we can convert into cubic meters with a typical unit conversion. So one m or 100 centimeters all cubed and this is equal to 100 25 multiplied by 10 to the power of negative six cubic meters. So let's use the pressure formula we found earlier to find the initial pressure of the gas. So P one is equal to N RT one divided by V. So let's plug the values into our calculator 0.25 moles N four, N recall that the ideal gas constant is 8.314 jewels per mole Calvin multiplied by 288 calvins all divided by a volume of 125 multiplied by 10 to the power of negative six cubic meters. And if we put that into a calculator, then we find an initial pressure of about 4. multiplied by 10 to the power of six pascals. So that's the initial pressure. Now let's find the final pressure. So P two is equal to N RT sub two divided by V. And note that the problem tells us that the final temperature T sub two is equal to 225 degrees Celsius. Which again, if we add 273 to convert to Kelvins, it's a temperature of 498 Kelvins. So let's plug the values into the P two equation. So once again, N is 0.25 moles, the ideal gas constant doesn't change 8.314 jewels per mole. Kelvin. Now we're plugging in for T sub 2, 498 Kelvins. And then we divide the entire thing by 100 25 multiplied by 10 to the power of negative six cubic meters. And if you plug that into a calculator, then we find a final pressure of 8.28 multiplied by 10 to the power of negative six pascals. So now we have found both the initial and final pressures of the gas. Now, the way the graph is set up is a little bit open ended because we're not given any values on the vertical pressure axis. So it's up to us to roughly estimate where those are. I'm going to kind of arbitrarily label one point on the vertical axis to correspond to 4.79, multiplied by 10 to the power of six pascals. And then at some distance above it maybe roughly around twice its height. I'll define that to be 8.28 multiplied by 10 to the power of six pascals. So now on the, on the horizontal axis where the volume is equal to 100 25 cubic centimeters. I will say that 0.1 is at the point where V equals 100 25 cubic centimeters and P is equal to 4.79, multiplied by 10, the power of six pascals. And this is 0.1. And I can kind of represent this using dotted lines to connect to the two points. So we can see this lies in the correct spot on both axes using these dotted lines. I can do the same thing with the upper point where the final pressure is. So the final pressure 0.2 is going to be the same at the same horizontal axis where the volume is equal to 100 25 cubic centimeters. But the pressure is equal to 8.28, multiplied by 10 to the power of six pascals. And this is 60.2. And so we can represent the process by drawing an arrow connecting the two points noting that the arrow should be pointing from 20.1 to 0.2 because we're starting off at a pressure of 4.79 mega pascals and then ending at a pressure of 8.28 mega pascals. So now this graph we've drawn here represents the process as described to us in the problem. And that's it for this problem. I hope this video helped you out. If it did. Lovely. If you still think you need more practice, please check out some of our other videos which will hopefully give you more experience with these types of problems, but that's all for now and I hope you all have a lovely day.

0.10 mol of argon gas is admitted to an evacuated 50 cm^3 container at 20°C. The gas then undergoes an isochoric heating to a temperature of 300°C. b. Show the process on a pV diagram. Include a proper scale on both axes.

Verified Solution

Key Concepts

Isochoric Process

Ideal Gas Law

pV Diagram