Question

. 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.

Answer 1

Do you know the ideal gas law?

Answer 2

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.

Answer 3

\[Pv = nRT \]

pressure times volume = the number of moles of gas times the gas constant times temperature

Answer 4

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.

Answer 5

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

\[\frac{ P _{1} V _{1}}{ V _{2} } = P _{2}\]

Answer 6

@Photon336 Can you please dumb down the answer please?

Answer 7

"shorten the answer"

Answer 8

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?

Answer 9

no :(

Answer 10

@Needhelpp101 for starters picture a container,

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

and they are going to hit the walls of the container

Answer 15

ok

Answer 16

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

Answer 17

yes

Answer 18

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

Answer 19

if we have something like this what do you see in these two figures

Answer 20

What can you say about V1 and V2?

Answer 21

hmm...

Answer 22

is V1 bigger than V2 or smaller?

Answer 23

smaller ?

Answer 24

V1 is bigger than V2 let me draw another one

Answer 25

k