Higher Physics 2011

Because you will never get all of the helium out of the cylinder, it will have some left inside it and the 0.08 is the volume of the cylinder - it says so at the top of the question.

Seems simple enough, but kind of stuck. Was wondering how to get the voltages VP and VQ in Question 25, part a) ii) in the 2010 paper.

Was unsure about the helium question too. It does say that that's the amount of gas the cylinder contains= 8.0*10^-2 m^3 and I'm assuming since they used a past tense (i.e. released) we are to assume that that amount of helium has already been used up to fill balloons, and hence deduct it from the total volume. Seems rather snide to be in a physics paper, though.

I think there has to be more to it than that? Surely?

Was unsure about the helium question too. It does say that that's the amount of gas the cylinder contains= 8.0*10^-2 m^3 and I'm assuming since they used a past tense (i.e. released) we are to assume that that amount of helium has already been used up to fill balloons, and hence deduct it from the total volume. Seems rather snide to be in a physics paper, though.

I think so too. But the only reason I could figure out is that they say so and so volume of gas is RELEASED. Hence adding a problem solving element in there. I hope for our sake that there's a REAL reason to it, that I'm missing. It'll be unnerving to know there are little tricks like that the examiners are playing on us.

P.S. Any idea on the Q. 25? Also, T = W/cos(x), correct?

Seems simple enough, but kind of stuck. Was wondering how to get the voltages VP and VQ in Question 25, part a) ii) in the 2010 paper.

T = W/cos(x), correct?

Vp = (1600/2400)x 12
Vq = (6000/10000)x 12
And then it it just Vp - Vq

And I don't know what you mean by T = W/cos(x)

But why is it 0.08? How do you know that?

No, it's not snide at all. Valid question and is in different problem collections. Just had a quick look and found this example. See Q5c in that exercise - it is same thing.

This type of problem is Higher outcome 2 (what was called problem solving at Standard Grade) according to the SQA.

Are you trying to rearrange the equation that finds the component of weight acting down a slope?

This is because once the pressure of the gas inside the cylinder matches that of
the filled baloons, there is no more pressure difference and no more baloons can be 'completely' filled with the Helium. Completely filled baloon is 0.020m^3 at 125KPa apparently!

I'm still not getting this. So when the cylinder starts to release gas the pressure goes from 750kPa to 125kPa and then stops and the total volume goes from 0.08 to 0.48. So why isn't there 0.48 in the balloons then?

The original volume of gas (cylinder only) at 750KPA expands to fill baloons completely and 0.08 m^3 remains in tank at 125KPa.
Think gas in cylinder -> gas in cylinder + gas in baloons.

Once the pressure has bottomed to 125KPA how could you get it out of the tank
into more baloons!

But why is it 0.08? Why not 0.07 or 0.09? Or is it just because that was there originally?

The cylinder is rigid made from steel presumably and can't change volume it is 0.08m^3 (the gas must fill the space available, unlike a liquid which will have a certain fixed volume) . The baloons can expand each to a volume 0.020m^3 as stated in the question.
You can't change the constants in a question to suit yourself, if the question says this is that etc then that is what it is wether you like it or not!