The Ideal Gas Law Derivations - Video Tutorials & Practice Problems

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The Ideal Gas Law Derivations are a convenient way to solve gas calculations involving 2 sets of the same variables.

Ideal Gas Law Derivations

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The Ideal Gas Law Derivations

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By rearranging the ideal gas law, we can derive new equations connected to pressure volume, moles and temperature. We're gonna say these derivations are required when we have variables with two sets of different values. So basically we'll be we'll be dealing with the question where the ideal gas laws in play and within the question they may give you two pressures and to 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, which is changing it a bit when we're dealing with two pressures or two volumes, two moles or to temperatures within any given question. Now that we've seen this, let's go on to our example question in the next video.

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example

The Ideal Gas Law Derivations Example 1

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4m

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in this example question. It says a sample of sulfur hexafluoride gas occupies 8.30 leaders at 202 degrees Celsius. Assuming that the pressure remains constant, what temperature degrees Celsius is needed to decrease the volume to 5. leaders. All right, so how do I know I'm dealing with an ideal gas derivation? Remember, we said that you're gonna have to derive a new formula any time off. Variable from the ideal gas law has two different values associated with it. If we look, we have two volumes and we have one temperature here and they're asking for a second temperature here. So we're dealing with two temperatures again. If you're dealing with two pressures, two volumes, two moles or to temperatures were dealing with an ideal gas derivation. So these are the steps we're gonna have to use to find her answer, begin by writing out the ideal gas law formula. So we're going to do that. We're gonna say PV equals N r. T. Now we're going to circle the variable in the ideal gas law formula that have two sets of different values. So we had again two volumes being discussed to 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 were discussing moles. Since they are constant, will have the same value, can also ignore it. Now we're gonna say algebraic Lee, 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 the over tea at this point, realize you're going to have to make these circle variables equal to the second set of identical variables in order to derive a new formula. And this is important. If the temperatures involved is involved in the calculation, it must use the S I unit of Calvin's. All right, so what did that last line mean? Well, we have the over t here, which is great, but remember, we have two volumes and two temperatures, so we'll make this one V one and t one and they will equal the second set of volume and temperature. So we've just arrived. The formula we're gonna have to utilize in order to find our answer. So now what we do is to just bring down the numbers that we had. So if we look, we had 8.30 leaders, 5.25 leaders. Since this leader has said first, we're gonna say it's V one. Since this temperature said first, this is t one, they're asking for a second temperature. So this is t two. Since this is the second volume being discussed, this is V two. All right, so I'm just gonna bring this formula up here. So v one over t one equals V two over T to plug in our numbers. So 8.3 leaders for V one and then 5. leaders for V two. Remember? We just said that your temperature and Kelvin that 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 2, 73 15. So it's gonna give us 4. 75 15. Kelvin and then t two is what we're looking for. We don't know what it is, so just solve for t two. We're gonna cross multiply these two, cross multiply these two. So when we do that, we're gonna have here 8.30 leaders times t two equals 4, 75 15 kelvin times 5. leaders the viable sides by 8.30 Leaders, leaders cancel out. And initially I'll have my temperature in Kelvin so t to initially equals 300.5 five. Kelvin, what we wanna we wanna answer in degree Celsius, So subtract out to 73.15. And when we do that, we're gonna get our degrees Celsius. So that comes out to when we work it out 17.40 degrees Celsius as our final answer.

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Problem

Problem

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

0.65 atm

B

0.83 atm

C

0.92 atm

D

1.6 atm

E

2.7 atm

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Problem

Problem

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?

A

0.176 L

B

0.217 L

C

0.246 L

D

0.361 L

E

1.28 L

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Problem

Problem

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

A

750 torr

B

1800 torr

C

2190 torr

D

2850 torr

E

3300 torr

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