Calculating Pressure Change in Gases: Hand Pump and Container Setup

[curiousjoe94]

Homework Statement

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.

 

[TSny]

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.

 

[curiousjoe94]

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.

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)?

 

[TSny]

Not quite.

Let P_o and V_o represent the pressure and volume of the air in the pump and P_f and V_f 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.

 

[curiousjoe94]

Not quite.

Let P_o and V_o represent the pressure and volume of the air in the pump and P_f and V_f 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.

how is this done?

I know that initial pressure is 101kPa

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

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

boyle's law is PV = constant

 

[Chestermiller]

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)?

 

[curiousjoe94]

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)?

Oh I see, so pressure would increase by 25% as well. So the new pressure would be 101kPa + 101*0.25kPa?

 

[TSny]

We are imagining that the air in the pump (volume = V_o and pressure P_o) is moved into the container (volume = V_f and pressure P_f). So, the volume of the air initially in the pump clearly changes from V_o to V_f . (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.]

 

[TSny]

Do you see the logic for why the number of moles in the container increases by 25%?

 

[curiousjoe94]

We are imagining that the air in the pump (volume = V_o and pressure P_o) is moved into the container (volume = V_f and pressure P_f). So, the volume of the air initially in the pump clearly changes from V_o to V_f . (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.]

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

 

[curiousjoe94]

Do you see the logic for why the number of moles in the container increases by 25%?

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%

 

[Chestermiller]

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?

 

[curiousjoe94]

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?

didn't Tsny say that this method can't be used because volume is dimensionless?

 

[Chestermiller]

didn't Tsny say that this method can't be used because volume is dimensionless?

No, he said that the ratio of the volumes is dimensionless. Volume has the dimension of length cubed.

 

[TSny]

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?

 

[curiousjoe94]

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?

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, 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?

 

[TSny]

"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.