A pressurized can of whipped cream has an internal pressure of 1.080 atm at 25°C. If it is placed in a freezer at -8°C, what is the new value for its internal pressure (in atmospheres)?

Answer 1

Hmmmm. What is the freezing temperature of the liquid in the can? I bet at -8, it is frozen. how much did it expand (after all, creame is mainly water and fat in an emulsion). So my point is, you can't assume the gas contents in the can have the same volume, due to the liquid expansion. But neglecting that reality.

P1/T1=P2/T2 temps in kelvins

Answer 2

So i tried (1.080)(-8)=(P)(265) and I got -30.67 but its no correct. What am I doing wrong?

Answer 3

A pressurized can of whipped cream has an internal pressure of 1.080 atm at 25°C. If it is placed in a freezer at -8°C, what is the new value for its internal pressure (in atmospheres)?

So i tried (1.080)(-8)=(P)(265) and I got -30.67 but its no correct. What am I doing wrong?

Answer 4

You didn't use kelvin for T.

Answer 5

1) Consider (P1V1)/T1 = (P2V2)/T2
Volume is constant so its P1/T1 = P2/T2
2) Convert C to K
3) Plug in

(Pressure and temperature have an inverse relationship unlike the direct relationship between pressure and volume.)