. If measurements of a gas are 50L and 300 kilopascals and then the gas is measured a second time and found to be 75L, describe what had to happen to the pressure (if temperature remained constant). Include which law supports this observation.

- Needhelpp101

- chestercat

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- Photon336

Do you know the ideal gas law?

- Needhelpp101

The ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behavior of many gases under many conditions, although it has several limitations. It was first stated by Émile Clapeyron in 1834 as a combination of Boyle's law and Charles's law.

- Photon336

\[Pv = nRT \]
pressure times volume = the number of moles of gas times the gas constant times temperature

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- Photon336

take a look at this formula, and I'll show you how to manipulate it b/c it's the only thing you'll need for the gases.
Now to use this equation we must do two things:
To study the change in any two of the variables we must keep one of those variables constant.

- Photon336

In our case the relationship is as follows:
we keep temperature constant, and when we do that well. we can ignore it in our formula so it becomes like this
\[pV = nR \]
now nR = the number of moles times the gas constant. we're asked to find the new pressure, when given a volume.
So it becomes because we're asked to find the new pressure.
\[P _{1}V _{1 } = P _{2}V _{2}\]
we assumed that the number of moles nR was constant so we omitted that. this shows us that pressure and volume are inversely proportional if i increase pressure by certain amount, then the volume must go down, if i increase volume by a certain amount i decrease my pressure. This makes sense because picture a sealed container, with some gas. if you decrease the volume, the gas particles have less space to move around, and when that happens they hit the walls of the container more frequently and the total pressure goes up. if you increase the volume, the pressure goes down, b/c more space available to the gas to move round less collisions with the walls of the container, at least that's how i understood it.
Now what are we solving for? that's P2
so
\[\frac{ P _{1} V _{1}}{ V _{2} } = P _{2}\]

- Needhelpp101

@Photon336 Can you please dumb down the answer please?

- Needhelpp101

"shorten the answer"

- Photon336

okay I can walk you through it. so for pressure and volume \[P _{1}V _{1} = P _{2}V _{2}\] if you start out with say some pressure and volume P1V1 and then you increase your volume your pressure goes down, that's what happened in your problem. but can you see why?

- Needhelpp101

no :(

- Photon336

@Needhelpp101 for starters picture a container, |dw:1437595340805:dw|

- Photon336

say we put a certain amount of gas in that container; what is that gas going to do?

- Needhelpp101

take up space......I'm dumb sorry thats why i'm on here :(

- Photon336

well the gas particles are going to move around, gases always move around

- Photon336

and they are going to hit the walls of the container

- Needhelpp101

ok

- Photon336

now think of the pressure as like how many times the gas particles hit the container make sense?

- Needhelpp101

yes

- Photon336

Now think of volume as like how much space is available for our gas to move around

- Photon336

|dw:1437595907989:dw| if we have something like this what do you see in these two figures

- Photon336

What can you say about V1 and V2?

- Needhelpp101

hmm...

- Photon336

is V1 bigger than V2 or smaller?

- Needhelpp101

smaller ?

- Photon336

V1 is bigger than V2 let me draw another one

- Needhelpp101

k

- Photon336

|dw:1437596265405:dw|

- Photon336

the way I drew the first one may have been confusing.

- Needhelpp101

k

- Photon336

what about the one I just drew?

- Needhelpp101

V1 is bigger than V2

- Photon336

Great

- Photon336

now we said a bigger volume = more space available to the gas

- Photon336

and pressure = the how many times the gas hits the walls of the container

- Photon336

if V1 has a greater volume than V2 then which one would the pressure be greater?

- Needhelpp101

V1?

- Photon336

another thing if there is less space for the gas to move around that means the gas hits the walls more so more pressure

- Needhelpp101

got ya

- Photon336

Gases are always moving

- Photon336

if there is less space for them to move they will hit the walls more

- Needhelpp101

ok

- Photon336

so from that figure with the squares which one would have the higher pressure and why?

- Needhelpp101

V2 because it's smaller

- Needhelpp101

We can use Boyle's Law, pressure is inversely related to volume when temperature is constant
P=kV 50=k300 k=50×300=15000 P=15000V
now plug in the new volume 75 L
P=1500075=200 kilopascals
Results:
Keeping temperature constant, when you increase the volume the pressure has to decrease. Boyle's law confirms this.
For a review of the gas laws
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch4/gaslaws3.html

- Photon336

Take a look at this

##### 1 Attachment

- Needhelpp101

k

- Photon336

That's correct, so do yo understand why that's the case based off what we said?

- Needhelpp101

yes :)

- Needhelpp101

Can you help me with one question?
You make the following measurements of an object – 22 kg and 42 m3. What would the object’s density be? Show your work for credit and include final units.

- Photon336

okay one last thing; so for now just think of pressure as
how many times the gas hits the walls of the container = pressure
volume = how much space gas has to move around.
increase volume, more space to move around so gas hits container less, pressure goes down.
decrease volume, less space to move around, hits containers more pressure goes up

- Photon336

can you put that question in another post? so you got the pressure volume stuff right?

- Needhelpp101

yep

- JTfan2000

Charles law - volume/temp are related and increasing

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