By rearranging the ideal gas law, we can derive new equations connected to pressure, volume, moles, and temperature. We're going to say these derivations are required when we have variables with two sets of different values. So basically, we'll be dealing with a question where the ideal gas law is in play. And within the question, they may give you two pressures and two temperatures, or two volumes with two moles. That's when we have to do one of these types of derivations. So just remember, we're still utilizing the ideal gas law, we're just changing it a bit when we're dealing with two pressures or two volumes, two moles, or two temperatures within any given question. Now that we've seen this, let's go on to our example question in the next video.
The Ideal Gas Law Derivations - Online Tutor, Practice Problems & Exam Prep
The Ideal Gas Law Derivations are a convenient way to solve gas calculations involving 2 sets of the same variables.
Ideal Gas Law Derivations
The Ideal Gas Law Derivations
Video transcript
The Ideal Gas Law Derivations Example 1
Video transcript
In this example question, it says a sample of sulfur hexachloride gas occupies 8.30 liters at 202 degrees Celsius. Assuming that the pressure remains constant, what temperature in degrees Celsius is needed to decrease the volume to 5.25 liters? Alright. So how do I know I'm dealing with an ideal gas derivation? Remember, we said that you're going to have to derive a new formula anytime a variable from the ideal gas law has 2 different values associated with it. If we look, we have 2 volumes, and we have one temperature here and they're asking for a second temperature here. So we're dealing with 2 temperatures. Again, if you're dealing with 2 pressures, 2 volumes, 2 moles, or 2 temperatures, we're dealing with an ideal gas derivation. So these are the steps we're going to have to use to find our answer. Begin by writing out the ideal gas law formula. So we're going to do that. We're going to say PV=nRT. Now we're going to circle the variables in the ideal gas law formula that have 2 sets of different values. So we had again 2 volumes being discussed, 2 temperatures being discussed. Next we're going to cross out the variables in the ideal gas law formula that are not discussed or remaining the same. So they weren't discussing pressure because it's being held the same, they weren't discussing moles. Since the R constant will have the same value, you can also ignore it. Now we're going to say algebraically move all the circled variables to the left side of the ideal gas law formula. So I need to move temperature to the other side, so I have to divide it out. So now my equation becomes V/T. At this point, realize you're going to have to make these circled variables equal to the second set of identical variables in order to derive a new formula. And this is important, if the temperature is involved in the calculation, it must use the SI unit of kelvins. Alright. So what did that last line mean? Well, we have V/T here which is great, but remember we have 2 volumes and 2 temperatures. So we'll make this V1 and T1, and they will equal the second set of volume and temperature. So we've just derived the formula we're going to have to utilize in order to find our answer. So now what we do is just bring down the numbers that we had. So if we look we had 8.30 liters, 5.25 liters. Since this liter is set first, we're going to say it's V1. Since this temperature is set first, this is T1. They're asking for a second temperature, so this is T2. Since this is the second volume being discussed, this is V2. Alright. So I'm just going to bring this formula up here. So V1 over T1 equals V2 over T2. Plug in our numbers, so 8.3 liters for V1 and then 5.25 liters for V2. Remember we just said that your temperature in Kelvin, your temperature has to be in Kelvin. Even though they want the answer in Celsius, when we're doing calculations we first have to convert the temperature into Kelvin. So we have 202 degrees Celsius plus 273.15. So it's going to give us 475.15 Kelvin. And then T2 is what we're looking for. We don't know what it is. So just solve for T2. We're going to cross multiply these 2, cross multiply these 2. So when we do that we're going to have here 8.30 liters times T2 equals 475.15 Kelvin times 5.25 liters. Divide both sides by 8.30 liters. Liters cancel out, and initially I'll have my temperature in Kelvin. So T2 initially equals 300.55 Kelvin. But we want our answer in degrees Celsius, so subtract out 273.15, and when we do that we're going to get our degrees Celsius. So that comes out to, when we work it out, 27.40 degrees Celsius as our final answer.
A sample of nitrogen dioxide gas at 130 ºC and 315 torr occupies a volume of 500 mL. What will the gas pressure be if the volume is reduced to 320 mL at 130 ºC?
A cylinder with a movable piston contains 0.615 moles of gas and has a volume of 295 mL. What will its volume be if 0.103 moles of gas escaped?
On most spray cans it is advised to never expose them to fire. A spray can is used until all that remains is the propellant gas, which has a pressure of 1350 torr at 25 ºC. If the can is then thrown into a fire at 455 ºC, what will be the pressure (in torr) in the can?
a) 750 torr
b) 1800 torr
c) 2190 torr
d) 2850 torr
e) 3300 torr