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OCR Physics G484 - June 2013 Unit 4 (OFFICIAL RETAKE THREAD)

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Original post by Muhammadjamal


Good, did you get 250 too?
Original post by EZCS
I dont know how to do the last question in queation 5. Does anyone know how to do?

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Use p=f/a do before and after and then minus one from the other

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Original post by EZCS
I dont know how to do the last question in queation 5. Does anyone know how to do?

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I got the answer but without seeing the paper I won't be able to remember.
Original post by niallh1995
Good, did you get 250 too?


246 (3sf) to be exact. I gt 1.5x10^7 in the end though

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fkh_95
the car was 1200kg... Did people remember to divide it by 4 (wheels)?
Original post by EZCS
I dont know how to do the last question in queation 5. Does anyone know how to do?

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I did change in pressure = force/area, the force being mg. Then rearrange for area
Original post by fkh_95
the car was 1200kg... Did people remember to divide it by 4 (wheels)?


Sh********t!!!! Didnt do that. Too happy i figured out what to do

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Reply 247
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theslav
For the very last question I added up the moles and the volumes, since I thought the most logical thing is that the volume of gas has also increased...in the end got an answer of 5x10^5 Pa
Original post by Muhammadjamal
246 (3sf) to be exact. I gt 1.5x10^7 in the end though

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yeah I got that, I always round to 2 sig figs though.
Original post by Muhammadjamal
Sh********t!!!! Didnt do that. Too happy i figured out what to do

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haha I did that but I was too happy that I realised I did force in KG XD
Original post by theslav
For the very last question I added up the moles and the volumes, since I thought the most logical thing is that the volume of gas has also increased...in the end got an answer of 5x10^5 Pa


You werent supposed to add the volumes. The extra gas was being put into the same container with constant volume of 0.05. The volume of the extra gas was given so you could work out how many extra mols were put in. Use the total number of mols to work out the new pressure

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theslav
Did it specify that the volume was of the box or was it of the gas inside it? Because I think that it said that volume of gas in the box was 0.05, so if you add more moles of gas then the volume of gas should increase, but the volume of the box is constant
Original post by theslav
Did it specify that the volume was of the box or was it of the gas inside it? Because I think that it said that volume of gas in the box was 0.05, so if you add more moles of gas then the volume of gas should increase, but the volume of the box is constant


I think it was about a gas cylinder and the volume was for that if someone can confirm pls

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Original post by fkh_95
the car was 1200kg... Did people remember to divide it by 4 (wheels)?


We had to divide by 4?! Jesus christ, bastard examiners...
Reply 254
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theslav
Original post by Ryan Gregg
We had to divide by 4?! Jesus christ, bastard examiners...


What if the car had three wheels?

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Original post by theslav
What if the car had three wheels?

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That is a fair point. I'm sure I would of realised it was insinuating working out the area per tyre instead of the area of all the tyres... God dammit.
Original post by Muhammadjamal
I think it was about a gas cylinder and the volume was for that if someone can confirm pls

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Original post by theslav
Did it specify that the volume was of the box or was it of the gas inside it? Because I think that it said that volume of gas in the box was 0.05, so if you add more moles of gas then the volume of gas should increase, but the volume of the box is constant


The volume of the cylinder would remain the same. All you're doing is increasing the number of moles and pressure.
Reply 257
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theslav
Original post by eggfriedrice
The volume of the cylinder would remain the same. All you're doing is increasing the number of moles and pressure.


2-3 marks lost for me then

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physicis
Original post by theslav
What if the car had three wheels?

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Dont worry guys you did not have to cosnider the number of wheels at all in the question. I specificaly though of this in the exam and reread the question all it asked for was the area in contact with the floor. No consideration of the number of wheels was necessary at all.

1200g=pressure1 x area1

1200g=pressure2 x area2

change area1-area2
Original post by physicis
Dont worry guys you did not have to cosnider the number of wheels at all in the question. I specificaly though of this in the exam and reread the question all it asked for was the area in contact with the floor. No consideration of the number of wheels was necessary at all.

1200g=pressure1 x area1

1200g=pressure2 x area2

change area1-area2


Thank god.

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