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### Course: AP®︎/College Chemistry > Unit 3

Lesson 4: Ideal gas law- The ideal gas law (PV = nRT)
- Worked example: Using the ideal gas law to calculate number of moles
- Worked example: Using the ideal gas law to calculate a change in volume
- Gas mixtures and partial pressures
- Worked example: Calculating partial pressures
- Ideal gas law

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# Worked example: Using the ideal gas law to calculate a change in volume

The ideal gas law can be used to describe a change in state for an ideal gas. In this video, we'll apply the ideal gas law to the initial and final states of a gas to see how changes in temperature and pressure affect the volume of the gas. Created by Sal Khan.

## Want to join the conversation?

- At6:15, how does it make sense that volume increases as the temperature decreases? Didn't he say in the last video that as temperature decreases, so does volume?(9 votes)
- I believe it is because the differential in temperature change (296K to 229K) is countered by the much larger change in pressure (765Torr to 6.51Torr) and therefor you still get a large increase in volume even with one of the variables counteracting that increase. P1(765Torr) * T2(229K) is a much larger number than P2(6.51Torr) * T1(296K).(9 votes)

- At1:19, why doesn't the # of moles change? Pressure and volume are changing in this question and isn't volume directly proportional to the # of moles (V=N)?

I have an additional question, so in this question-T2 (final temperature) is referring to the temperature inside the balloon?(2 votes)- How do you suppose the number of moles could change though? The gas is inside the ballon, you should be assuming n is constant.

V=n isn’t true, they’re proportional yes, but not equal. The proportionality depends on the temperature and pressure like the ideal gas law shows.

The temperature refers to the gas molecules inside the ballon yes.(15 votes)

- I tried to answer the example before I continued to watch the solution, and what I did was I tried to find the value of n first by using the given initial values. After obtaining, I solved for the value of V2 by dividing P2 to nRT2 but when I watched the solution, my answer is different to Sal's answer. I would like to know what's the error in my procedure.(4 votes)
- That's another way to calculate the final volume which works. For your method to work you need to use 'R' and you have to make sure you use the correct version of 'R' to line up with the rest of the units in the problem. Pressure is in torr, volume is in liters, and temperature is in kelvin which means your value of the universal gas constant should have those same units. Which means you want to use 62.364 with units of L*torr*K^(-)*mol^(-). That was most likely your mistake, that you used the incorrect version of 'R'. Hope that helps.(4 votes)

- I'm stuck at the reasoning that the instructor initially used to formulate the equation.

If we're solving for Volume then wouldn't we divide P from both sides of the Ideal Gas Law?

PV=nRT ---> V = (nRT) / P

I understand the process of solving the equation once it has been constructed, but I'm lost as to the reasoning behind formulating it that way to begin with.

Hope someone can shed some light of this for me!

Best Regards,

Alex(1 vote)- So if we began with the ideal gas law and wanted to solve for volume, that would indeed be the equation we would use: V = (nRT)/P. However this use with just using this equation is that we don't just want to calculate volume at a single state, we want to calculate the volume at a new second state.

The problem with you equation is that it requires us to know the temperature, pressure, and moles of the gas to solve for the volume. However the problem gives no information about the number of moles, for either the initial or final state.

The idea with Sal's equation is that we rearrange the ideal gas law to that all the changing variables are on one side of the equation and the unchanging variables (the constants essentially) are on the other side. In this problem we don't know what the exact value of n is, but we know this it is unchanging is not gaining or losing gas so n is a constant in addition to R. So Sal's equation: PV/T = nR, has the changing variables on the left and the constants in the right. And this equation is true for both the initial and final states so we essentially have two versions of the same equation, just with different subscripts to denote the state. So if these two equations are equal to the same constant, then through the transitive property these two equations are equal to each other. Then through some algebra rearrangements we arrive at an equation that solves for the final volume.

Hope that helps.(6 votes)

- At5:35weren't you supposed to multiply the 6.51 Torr by the 296 K?(3 votes)
- for the scientific notation, could you not just write 168 * 10^3? what is the point of the two decimals in 1.68?(2 votes)
- Mathematically, 168x10^(3) would be equivalent to 1.68x10^(5). So, technically both would be correct answers. And they would both be in scientific notation since they’re expressed using powers of ten.

Restricting the significant portion of the number, known as the mantissa, to values between 1 – 10 normalizes the number. Therefore, using scientific notation with a normalized mantissa is referred to as normalized scientific notation. Normalized scientific notation is the most common way to express scientific notation. This form allows easy comparison of numbers: numbers with bigger exponents are larger than those with smaller exponents, and subtraction of exponents gives an estimate of the number of orders of magnitude separating the numbers. So, while 168x10^(3) would be scientific notation, 1.68x10^(5) would be normalized scientific notation and would be the more widely accepted answer of the two.

Interestingly there’s another scientific notation known as engineering notation. This one allows the mantissa take on values between 1-1000, but only allows exponents in multiples of 3. Engineering notation allows the numbers to explicitly match their corresponding SI prefixes (kilo, centi, milli, etc.).

Hope that helps.(3 votes)

- why did he count 5 places for the scientific notation superscript? I counted 6.(2 votes)
- How does the volume
**inside**the balloon change when it gets to a higher / lower temperature?(1 vote)- As a clarification, the temperatures used in these gas problems apply to the temperature within the container such as a balloon. When we change the temperature, either as an increase or decrease, this causes a corresponding change in the kinetic energy of the gas particles. Kinetic energy is the energy of motion, so this will also change the gas molecules’ speeds. A different speed for the particles means they collide with the inside of the container with a different force, which we interpret as pressure. A different pressure therefore causes a change in volume as the gas particles push harder or weaker against the container’s surface (causing a contraction or expansion of the container).

Hope that helps.(2 votes)

- If the number of moles remains constant, what exactly is filling up the balloon to make it bigger?(1 vote)
- The balloon is not gaining any additional air. The moles being constant means that the air inside the balloon is constant. The reason the balloon increases in volume is due to the change in atmospheric pressure.

The balloon has an internal pressure due to the gas inside pushing out against the inside of the balloon. It also has an external pressure due to the atmosphere’s air pushing against the outside of the balloon. Before the balloon is launched, both of these forces are equal and opposite resulting in its initial, stable volume. After it is launched, it ascends into the atmosphere where the atmospheric pressure gradually decreases. This is because as you move higher in the atmosphere, the amount of air particles decreases resulting in less collisions with surfaces (which is what pressure is). If the atmospheric pressure decreases, but the internal pressure of the balloon remains constant, then there is now a greater force pushing out compared to before it was launched. The internal pressure meeting less resistance from the atmospheric pressure causes the volume to increase.

Hope that helps.(3 votes)

- What If we are given a question with only volume and temperature at two different times?(2 votes)
- Well what are you trying to calculate? If you have an initial volume and temperature then undergo a change you now have a final volume and temperature. Then presumable you're either trying to calculate pressure or the number of moles, but which one?(1 vote)

## Video transcript

- [Instructor] We're told
that a weather balloon containing 1.85 times
10 to the third liters of helium gas at 23 degrees
Celsius and 765 Torr is launched into the atmosphere. The balloon travels for two hours before bursting at an
altitude of 32 kilometers, where the temperature is
negative 44 degrees Celsius and the pressure is 6.51 Torr. What is the volume of the
balloon just before it bursts? So pause this video and see
if you can figure that out. All right, so you might
already have an intuitive sense that this has something to
do with the ideal gas law, because they're giving
us a bunch of pressures, volumes, and temperatures, and the ideal gas law deals with that, it tells us that pressure times volume is equal to the number of moles times the ideal gas
constant times temperature. Now, what's different about this example is that they aren't just giving us several of these variables and asking us to solve one of them, they're talking about
these variables changing, and how that might affect other variables. And so one way to think
about it is if we divide both sides by T, you get
PV over T is equal to NR. And in this example, as this balloon goes to higher and higher altitudes, the number of moles does not change, and the ideal gas
constant does not change. So one way to think about it is that PV over T has to be constant. So our volume and our
temperature could change, but because this whole
expression on the left has to be constant, that could
then determine our pressure. Or another way to think about it, you could say your starting pressure times your starting volume
over your starting temperature is going to be equal
to the number of moles times the ideal gas constant, which also needs to be
equal to your pressure right before it bursts times the volume right before it bursts,
divided by the temperature right before it bursts, or you could just say
that P one times V one over T one is equal to P
two times V two over T two. And so what are these different variables? Well, let's first think about P one, so pressure at time one
is what, it's 765 Torr. 765 Torr, and what's P two? That's the pressure just before it burst, and they tell us it's 6.51
Torr, much lower pressure, which makes intuitive sense,
we're at a higher altitude. 6.51 Torr, now, what is V one? Well, they tell us that right over there, that is 1.85 times 10 to the third liters. Now, what is V two? Well, that's what they
want us to figure out, what is the volume of the
balloon just before it bursts? So I'll put a little question mark there. And then, last but not
least, what is T one? Well, they tell us the
starting temperature is at 23 degrees Celsius, but you have to think on
more of an absolute scale, and deal with temperatures
in terms of Kelvin, so to convert 23 degrees
Celsius into Kelvin, you have to add 273, so this
is going to be 296 Kelvin, and then what is T two? Well, T two is negative
44 degrees Celsius, if we add 273 to that, let's see, that's going to be, if we
subtract, it's going to be in my head, 229 Kelvin. And so we have everything we need in order to solve for V two, in fact, we can solve for V two before we even put in these numbers, if we multiply both sides of this equation times T two over P two, and the reason why I'm
multiplying it times this is so that this cancels with
this, this cancels with that, so I have just V two
on the right-hand side. Of course, I have to
do that on both sides. T two over P two, I am going to get, and I'll now color code it, I'm going to get that T two times P one times V one, over P
two times T one, T one, is equal to V two, V two. So we just have to
calculate this right now, let me give myself a
little bit more real estate with which to do it, and so we could write that
V two is equal to T two, which is 229 Kelvin, times
P one, which is 765 Torr, times V one, which is 1.85
times 10 to the third liters, all of that over P two,
which is 6.51 Torr, times T one, which is 296 Kelvin, and we can confirm that
the units work out, Torr cancels with Torr,
Kelvin cancels with Kelvin, so we're just gonna have a
bunch of numbers, a calculation, and the units we're left with is liters, which is good, because
that is what we care about when we care about volume. So this is going to be
equal to 229 times 765, times 1.85 times 10 to the third, divided by 6.51, divided by 296 is equal to this business right over here. Let's see, we have three
significant digits here, three significant digits here, three here, three here, and three there, so our answer's going to have
three significant figures, so it's going to be, if we round, it's gonna be 168,000, and
so we could just write that as 168,000 liters, or
if we wanna write that in scientific notation,
we could write that as 1.68 times 10 to the
one, two, three, four, five, so let me write it that way, so this is going to be equal to 1.68 times 10 to the fifth liters. And I always like to do
a nice intuition check, does that makes sense? So our starting volume was 1,850 liters, and then our volume got a lot larger, because we're going to
a much higher altitude, and that does make intuitive sense to me.