What temperature (in °C) did an ideal gas shift to if it was initially at -16.0 °C at 4.62 atm and 35.0 L and the pressure was changed to 3.25 atm and the volume changed to 40.0 L?

Answer 1

Given:

Initial temperature T1 = -16.0ºC + 273 = 257K

Initial pressure P1 = 4.62 atm

Initial volume = V1 = 35.0 L

Final pressure = P2 = 3.25 atm

Final volume = V2 = 40.0 L

Asked for:

Find the final temperature = T2

Procedure:

Use the combined gas law: P1V1/T1 = P2V2/T2 and solve for T2. Then convert back to ºC

(4.62 atm)(35.0 L) / (257K) = (3.25 atm)(40.0 L) / T2

T2 = (275K)(3.25 atm)(40.0 L) / (4.62 atm)(35.0 L)

T2 = 221K

221K - 273 = -51.9ºC

(please be sure to check the math)