A gas that has a volume of 28 liters, a temperature of 42°C, and an unknown pressure has its volume increased to 49 liters and its temperature decreased to 27°C. If the pressure is measured after the change at 4.0 atm, what was the original pressure of the gas?

7.35atm

Explanation:

Data obtained from the question include:

V1 = 28L

T1 = 42°C = 42 + 273 = 315K

P1 =?

V2 = 49L

T2 = 27°C = 27 + 273 = 300K

P2 = 4atm

Using P1V1/T1 = P2V2/T2, the original pressure can be obtained as follows:

P1V1/T1 = P2V2/T2

P1 x 28/315 = 4 x 49/300

Cross multiply to express in linear form

P1 x 28 x 300 = 315 x 4 x 49

Divide both side by 28 x 300

P1 = (315 x 4 x 49) /(28 x 300)

P1 = 7.35atm

Therefore, the original pressure is 7.35atm

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