A soft drink is made by dissolving CO2 at 3.00 atm in a flavored solution and sealing the solution in an aluminum can at 20oC. What volume of CO is released when a 355-mL can is opened to 1.00 atm at 20oC and all CO is allowed to escape

Volume is 1.065L

Explanation:

Hello,

We can easily solve this problem by using general gas equation.

PV / T = K

P1V1/T1 = P2V2/T2

Data;

P1 = 3.0atm

P2 = 1.0atm

T1 = 20°C = (20 + 273.15)K = 293.15K

T2 = 20°C = (20 + 273.15)K = 293.15K

V1 = 355mL = 0.355L

V2 = ?

From the data given, we can substitute it into the equation,

(P1 × V1) / T1 = (P2 × V2) / T2

(3.0 × 0.355) / 293.15 = (1.0 × V2) / 293.15

1.065 = 1.0V2

Divide both sides by 1.0

V2 = 1.065L

The volume of CO₂ released is 1.065L

it would be a gas formation