Once more with feeling! another ideal gas question.

1. Feb 7, 2006

QueenFisher

[SOLVED] once more with feeling! another ideal gas question.

the pressure inside a sealed can which is kept in a deep freeze at -23C is 0.8x10^5 N/m^2. If the can is placed in water of temperature 27C:

calculate the increase in pressure inside the can.

initially, pressure/temperature = 0.8x10^5 / 273-23
=320
afterwards, pressure 2 divided by (273+27) must equal 320 also. this gives pressure 2 as 96000Pa
increase in pressure = 0.8x10^5 - 96000
= -16000
but if the temperature increases, surely the pressure has to increase? since they are directly proportional??
or have i taken them away the wrong way round?

2. Feb 7, 2006

Galileo

The pressure HAS increased. 96,000 is larger than 80,000 after all.

3. Feb 7, 2006

vaishakh

I think you need much more information to solve this. We have to find the finla temperature of the air in the can to solve the following.

4. Feb 7, 2006

QueenFisher

so i guess i was taking them away the wrong way around. :yuck: man i'm stupid.

5. Feb 7, 2006

vaishakh

Or I think it has to be assumed that the temperature of the system becomes 300K when put in water nadwas 250K initially. Now use the same relation as you said - P1/T1 = P2/T2.
You have mistook the sign I would say. You made a silly mistake. Your problem solving is okay.

6. Feb 7, 2006

QueenFisher

yeah that was the equation i was used, just in some weird convoluted form