# Homework Help: 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