# Ideal Gas Law: What is the new pressure based on factors?

• ptownbro
In summary: P)P = [ (4 x 0.0821 x 2) / 2 ] x (nT / V)P = 0.0176 nT / VIn summary, our daughter has a homework problem that we need assistance in confirming if we've understood and completed correctly. The Volume goes from 2.00 Liters to 7.00 Liters, the temperature changes from 800 kelvin to 150 kelvin, and the number moles of gas is quadrupled on a gas sample initially at 1.20 atm. What is the new pressure? The Attempt at a Solution We assumed that since moles was used here, we need to use the "PV = nRT
ptownbro
My daughter has the following homework problem we need assistance in confirming if we've understood and completed correctly.

## Homework Statement

The Volume goes from 2.00 Liters to 7.00 Liters, the temperature changes from 800 kelvin to 150 kelvin, and the number moles of gas is quadrupled on a gas sample initially at 1.20 atm. What is the new pressure?

## Homework Equations

PV = nRT, where P = pressure, V = volume, n = moles, R = a given number at 0.0821, and T = temperature.

## The Attempt at a Solution

We assumed that since moles was used here, we need to use the "PV = nRT" version of the ideal gas equation versus the "P1V1 / T1 = P2V2 / T2" version. Given that, we also assumed we'd have to apply the factor of the changes in each applicable variable.

Volume factor = 7 / 2 = 3.5
Temperature factor = 150 / 800 = 0.1875
Moles factor = 4

Therefore, mathematically, logic stands you can apply the product of those factors to the given starting pressure amount of 1.20 atm to get the new pressure amount.

PV = nRT
P = nRT / V
P = [ (4 n) x (0.0821) x (0.1875 T) ] / (3.5 V)
P = [ (4 x 0.0821 x 0.1875) / 3.5 ] x (nT / V)
P = 0.0176 nT / V

Your factor is therefore 0.0176 which can be applied to the original pressure to get your new pressure.

P = 1.2 atm x 0.0176
P = 0.0211 atm

Can you please confirm if our logic and answer is correct and if not where we went wrong?

Thanks!

You can actually use the following:
(P1xV1)/(n1xT1)=((P2xV2)/(n2xT2)
it accounts for the change in moles also.

If you want me to show you why this is, i can show you.

Paul I. said:
(P1xV1)/(n1xT1)=((P2xV2)/(n2xT2)

That's interesting. Didn't know you could do that. Yes, be interested in seeing how you got to there.

However, I don't think I can use it here because I don't know what the original moles are. I just know they quadrupled.

Oh ok. Yes, then in that case, since you know the initial temperature, pressure, and volume. You can then solve for the amount of moles. Once you have the answer for the amount of initial moles, you can multiply that by 4, then solve for the final pressure. That's how i would go about it.

Paul I. said:
Oh ok. Yes, then in that case, since you know the initial temperature, pressure, and volume. You can then solve for the amount of moles. Once you have the answer for the amount of initial moles, you can multiply that by 4, then solve for the final pressure. That's how i would go about it.

Would you mind demonstrating?

I solved it in this image:
Hopefully it's what you're looking for!

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ptownbro
ptownbro said:
Given that, we also assumed we'd have to apply the factor of the changes in each applicable variable.

While technically that could be possible, it didn't yield the correct answer (but I have not checked why), and it is the most complicated way of dealing with the problem I can think of.

Why not split it into two simple problems:

1. Calculate number of moles present in the initial conditions.

2. Multiply by 4.

3. Calculate new pressure knowing n, R, T and V.

ptownbro
Ah. Makes total sense. Don't know why I didn't think of that approach. Should have kept it simple.

Still... not sure why mathematically our approach doesn't work since logically you should be able to apply the product of the ratios/factors.

For example, here's another problem where we used basically the same principals. And we THINK we got it right.

Question:
The pressure doubles, the absolute temperate triples, and the number of moles becomes 1/6 as great on a 1.2 L nitrogen sample. What is the new gas volume?

PV = nRT
V = nRT / P
V = [ (1/6 n) x (0.0821) x (3 T) ] / (2 P)
V = [ (1/6 x 0.0821 x 3) / 2 ] x (nT / P)
V = 0.02025 nT / P
V = 1.2 L x 0.02025
V = 0.02463 L

In this case, you aren't given any values except factors. So, not sure how we could use your approach.

If I can infringe on a little more of your time... how would you go about that one?

Thanks

Borek said:
Why not split it into two simple problems: 1. Calculate number of moles present in the initial conditions. 2. Multiply by 4. 3. Calculate new pressure knowing n, R, T and V.

Yes. This seems like same suggestion as Paul. Thanks to you too!

Yet... still not sure how you would apply it to a similar question where we used that answer. See my other reply to Paul.

Wait! I think I figured it out!

I don't think I was supposed to include the given amount for "R" at 0.0821 in my approach to figure out the factor amount.

PV = nRT
P = nRT / V
P = [ (4 n) x (R) x (0.1875 T) ] / (3.5 V) <----- Used that instead of: [ (4 n) x (0.0821) x (0.1875 T) ] / (3.5 V)
P = [ (4 x 0.1875) / 3.5 ] x (nRT / V)
P = 0.2143 nRT / V
P = 1.2 atm x 0.2143
P = 0.2571 atm

The same answer that Paul got

This makes sense because

P = nRT / V

So the factor would need to be applied to "nRT / V" and not "nT / V" as I did in my original.

i hope it's not too messy. i used the equation i suggested before. then i just ended up using the coefficients and canceled out the P, n, and T. hope this makes sense.

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ptownbro
0.257 atm is what I got too.

Sorry, no time for more ATM.

Edit: and I can confirm 0.3 L for the second problem.

ptownbro
here's how my equation came up to be. this is how i learned it.

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ptownbro
Seems like you got it! Glad you are able to understand it! I'll say goodnight to you and hope to be able to help you in the future!

Awesome. Thanks to both.

My adjusted approach (which now I know and admit is likely overly complicated) also gets 0.3 L.

PV = nRT
V = nRT / P
V = [ (1/6 n) x (R) x (3 T) ] / (2 P)
V = [ (1/6 x 3) / 2 ] x (nRT / P)
V = 0.25 nRT / P
V = 1.2 L x 0.25
V = 0.3 L

I think we get it now.

Thanks again!

Going to bed now. =)

Put to your daughter the way I would have done it, because I don't like to see young students struggling hard with easy things that they will then forget because they seem complicated. (And OK you do have to know some semi-elementary things for this, dependence on mass density, Boyle's Law, Charles' Laws, meaning of absolute temperature. And OK,Borek has said more or less the same thing,)

I'd just say: the pressure was so much. Then into the same volume I have pushed 4× as much stuff, mass so pressure is now

I have then let it expand to 7/2 × what it was, in other words made the density 2/7 × what it was

2/7 ×

then I have cooled it to 150/800 of what it was so pressure becomes

150/800 ×

what is was after the last operations: in all its pressure has become

4× 2/7 × 150/800 × than the 1.2 atm it originally was I.e. 0.257 atm , in agreement with others.

Simple proportions come into so many quantitative chemistry questions. Hope this helps your daughter.

Not to disdain the equations in the end, have to know about R etc., but helps make plain and less intimidating what they mean.

Last edited:
Borek
Borek said:

Will save you taking hard to read pictures :)
Yes thank you!I knew it was possible to write equations but i didn't know how because it looked highly complicated when ifirst tried! I'll be able to practice and make sure i know what I'm doing now!

Here is how I derived the equation to solve for final pressure. This show's you where your idea of multiplying by ratios could work if done properly. You don't need to calculate the moles in the beginning, then multiply by 4, then stick new number in...the moles cancel out as you'll see...
plug ur numbers in and see if this gives you the right solution; and remember to watch your units

## 1. What is the ideal gas law?

The ideal gas law is a mathematical equation that relates the pressure, volume, temperature, and number of moles of a gas. It is expressed as PV = nRT, where P is pressure, V is volume, n is number of moles, R is the gas constant, and T is temperature.

## 2. How is the new pressure calculated based on factors?

The new pressure can be calculated by using the ideal gas law equation and rearranging it to solve for pressure. This involves plugging in the known values for volume, temperature, and number of moles, and solving for pressure.

## 3. What factors affect the pressure in the ideal gas law?

The pressure in the ideal gas law is affected by the volume, temperature, and number of moles of the gas. As these factors change, the pressure will also change according to the ideal gas law equation.

## 4. Can the ideal gas law be used for all gases?

The ideal gas law is a theoretical equation that applies to all gases, but it is most accurate for gases at low pressures and high temperatures. At high pressures and low temperatures, the behavior of real gases deviates from the ideal gas law.

## 5. How does the ideal gas law relate to the behavior of gases in real life?

The ideal gas law serves as a useful approximation for the behavior of gases in real life, but it is not a perfect model. Real gases may exhibit deviations from the ideal gas law at certain conditions, such as high pressures or low temperatures, due to factors like intermolecular forces and the volume occupied by gas molecules.

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