1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Simple gas/pressure question

  1. Jun 15, 2012 #1
    1. The problem statement, all variables and given/known data

    A hand pump of volume 2.0 x 10-4 m3 is used to force air through a valve into a container of volume 8.0 x 10-4 m3 which contains air at an initial pressure of 101kPa. Calculate the pressure of the air in the container after one stroke of the pump, assuming the temperature if unchanged.




    2. The attempt at a solution


    Firstly bear in mind that this is the first question of the Gases section in my A2 physics textbook, so I have a hunch that this question should be quite simple, as well as the fact that only Boyle's law has been covered so far in this section.

    So anyway, I assume that Boyle's law is involved: PV = k (where P = pressure, V = volume, k = constant). I started by fiinding the constant by subbing in the initial pressure (of the container) and its volume to get constant k = 80.8

    I then subbed the constant value I've just found, and the volume of 8.0 x 10-4 m3 minus 2.0 x 10-4 m3 into the same PV = k equation. I figured that if air is being pumped in, then the container's volume must be reduced as the pumped in air takes up space, hence why I put volume as: 8.0 x 10-4 m3 minus 2.0 x 10-4 m3.

    However I get pressure as 134kPa. The answer at the back have it as 126kPa.
     
  2. jcsd
  3. Jun 15, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Boyle's law (PV = k) is only true if both the temperature and amount of ideal gas remains constant. In this problem, you are changing the amount of gas in the container.

    Here's one way to think about it……

    The final pressure of the container will be the initial pressure of the container plus the pressure that the added air would produce if it alone occupied the container. (This statement is valid for ideal gases only!)

    Try to use Boyle’s law to find the final pressure that the air in the pump would have if it were to occupy the volume of the container. Then use this result to find the final total pressure in the container.
     
  4. Jun 15, 2012 #3
    For this, do I just find the ratio between the volumes of the hand pump and the container, and then equate this ratio to the initial pressure (before the pump of air was put into the container)?
     
  5. Jun 15, 2012 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Not quite.

    Let Po and Vo represent the pressure and volume of the air in the pump and Pf and Vf represent the pressure and volume of this same amount of air if it occupied the volume of the container.

    Carefully use Boyle's law to relate the initial and final pressures and volumes. You said that the ratio of the volumes would equal the initial pressure. That can't be true because a ratio of volumes is dimensionless, whereas pressure does have dimensions.
     
  6. Jun 15, 2012 #5
    how is this done?

    I know that initial pressure is 101kPa

    and like you said final pressure would be Pf + 101kPa

    initial and final volume would surely be the same, right?

    boyle's law is PV = constant
     
  7. Jun 15, 2012 #6
    According to the ideal gas law,

    PV = nRT

    If the volume and temperature of the gas in the container are constant, how does the pressure in the container change if the number of moles of gas in the container is increased by 25% (assuming that the starting temperature and pressure of the gas in the pump was the same as in the container)?
     
  8. Jun 15, 2012 #7
    Oh I see, so pressure would increase by 25% as well. So the new pressure would be 101kPa + 101*0.25kPa?
     
  9. Jun 15, 2012 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member

    We are imagining that the air in the pump (volume = Vo and pressure Po) is moved into the container (volume = Vf and pressure Pf). So, the volume of the air initially in the pump clearly changes from Vo to Vf . (But you know what these volumes are.)

    The pressure of the air in the pump will also change as it expands from the volume of the pump to the volume of the container. (By the way, what is the value of the initial pressure of the air in the pump?)

    So, both the pressure and volume change for the air in the pump as it moves from the pump into the container. But Bolye's law gives you a way to relate the initial pressure and volume to the final pressure and volume. Can you see what the relation is?

    [Chestermiller is giving you another way to approach the problem using the full ideal gas law. I was trying to stick with Boyle's law. But, whatever works for you is good.]
     
  10. Jun 15, 2012 #9

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Do you see the logic for why the number of moles in the container increases by 25%?
     
  11. Jun 15, 2012 #10
    I'm not told the initial pressure of the air in the pump. I guess I'm going to have to assume that its the same as the container.

    I guess chestermiller's use of full ideal gas law works, but I find it strange because this question is on the first page of this gas section on the textbook where only boyle's law is introduced (full ideal gas law is on the pages after).

    I'm still not sure how I could use boyle's law to relate the initial and final pressure. The constant stays the same, yes?

    So are you suggesting this:

    (PV = constant)

    initial pressure of air in pump * initial volume of the pump = constant

    101kPa * 2.0 x 10-4 m3 = 20.2


    final pressure of air from pump in container * final volume of air from pump in container = constant

    P * 8.0 x 10-4 m3 = 20.2

    Therefore P = 25250Pa (0.252kPa)



    So 101kPa + 0.252kPa = 126.2kPa
     
  12. Jun 15, 2012 #11
    judging from the full ideal gas law he's given (PV = nRT)

    He said that the number of moles of the gas (n) in the container increases
    by 25% (not entirely sure how he worked this out)

    Therefore if V stays the same, then pressure must increase by 25%
     
  13. Jun 15, 2012 #12
    Yes. As the other responders indicated, you can also use Boyle's law to solve this problem. Initially, what is the total volume of gas contained in the pump plus the container? Assuming that initially, the pressure and temperature within the pump is equal to the pressure and temperature in the container, and the gas is compressed such that its new total volume is only that of the container. what is the ratio of the new total volume of the gas to the initial volume of the gas? How does this affect the ratio of the initial to the final pressure?
     
  14. Jun 15, 2012 #13
    didn't Tsny say that this method can't be used because volume is dimensionless?
     
  15. Jun 15, 2012 #14
    No, he said that the ratio of the volumes is dimensionless. Volume has the dimension of length cubed.
     
  16. Jun 15, 2012 #15

    TSny

    User Avatar
    Homework Helper
    Gold Member

    I didn't say volume is dimensionless. I said that a ratio of two volumes is dimensionless.

    Note that Chestermiller is not saying that the ratio of volumes equals the initial pressure (as you had proposed). He's saying that the the ratio of volumes is somehow related to the ratio of pressures. Ratios of volumes and ratios of pressures are both dimensionless.

    You need to figure out how to relate the ratio of initial and final pressures to the ratio of the initial and final volumes using Boyle's law.

    Chestermiller has a nice way of looking at the problem by considering the total amount of air in the entire system (pump and container). The total amount of air in the system doesn't change, nor does the temperature. So, you can apply Boyle's law to the total amount of air as it goes from atmospheric pressure (101 kPa) to the final pressure while the volume goes from the total initial volume (pump plus container) to the final volume (container alone).

    You don't have to think in terms of ratios. How does PV at the start relate to PV at the end?
     
  17. Jun 15, 2012 #16
    PV at the start is going to be the same as PV at the end, right?

    Going back to what I said about ratio of volumes, what I originally meant to say was that the ratio of volumes to be equal to the ratio of pressures, heres what I mean:

    volume of hand pump / volume of container

    equals

    additional pressure from the air being pumped in / initial pressure of the container


    would it be right to look at it this way?
     
  18. Jun 15, 2012 #17

    TSny

    User Avatar
    Homework Helper
    Gold Member

    "PV at the start is going to be the same as PV at the end, right?"

    As long as you are considering a fixed amount of air (at a fixed temperature).

    "Going back to what I said about ratio of volumes, what I originally meant to say was that the ratio of volumes to be equal to the ratio of pressures, here's what I mean:

    volume of hand pump / volume of container

    equals

    additional pressure from the air being pumped in / initial pressure of the container

    would it be right to look at it this way?"


    Here, you are considering just the air that was originally in the pump, right? OK. But, then you should be thinking:

    volume of hand pump / volume of container

    equals

    additional pressure from the air being pumped in / initial pressure of the air in pump

    Now, the initial pressure of the container just happens to be equal to the initial pressure in the pump. So, your way of writing it would be technically correct, but it makes me wonder if you are completely understanding the use of Boyle's law here.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Simple gas/pressure question
Loading...