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

the correct answer would be b. amelia, because the particals of gas move really fast (charged particals) and gas also has a lot of room (particals move independently) so the only answer that has both of those is amelia's.