Question

Zippy decides he needs a drink to wash down the cold Pring Pongs. There are two 355 mL cans of Kookie Cola in the mini-bar.
The Kookie Cola is carbonated and the refrigerated can is under 36.5 psi pressure even at 2.4°C. Zippy drinks one can and
accidentally leaves the other one on the dresser. If the pressure in the can rises to 38.9 psi, what is the temperature in Zippy's
hotel room at this point? Assume the solubility of the carbon dioxide is not affected by this temperature change.

Answer 1

20.7 Degree celsius is the temperature of Zippy's room.

Explanation:

The initial conditions of the kookie cola will be depicted by P1,V1 and T1.

P1= 36.5 psi or 2.4816 atm

V1= 355ml or 0.35 L

T1= 275.5K

The final conditions are given P2,V2 and T2

P2= 38.9 psi or 2.6469 atm

V2 = 355 ml or 0.35 L

also the solubility of the carbon dioxide does not change with temperature.

Applying the formula

P1VI/T1=P2V2/T2

since volume does not change and remains constant.

It can be written as:

P1/T1=P2/T2

= 2.4816/275.5=2.6469/T2

T2= 293.85 K