A gas that has a volume of 28 liters, a temperature of 45 °C, and an unknown pressure has its volume increased

to 34 liters and its temperature decreased to 35 °C. If I measure the pressure after the change to be 2.0 atm,

what was the original pressure of the gas?

2.5 atm

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15) K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15) K = 308.15k

P2 = 2.0atm

P1 = ?

From general gas equation,

(P1 * V1) / T1 = (P2 * V2) / T2

P2 * V2 * T1 = P1 * V1 * T2

P1 = (P2 * V2 * T1) / (V1 * T2)

P1 = (2.0 * 34 * 318.15) / (28 * 308.15)

P1 = 21634.5 / 8628.2

P1 = 2.5 atm

The initial pressure of the gas is 2.5atm