Question

A gas that has a volume of 15 liters, a temperature of 51°C, and an unknown pressure has its volume increased to 25 liters and its temperature decreased to 41°C. If I measure the pressure after the change to be 3.9 atm, what was the original pressure of the gas?

Answer 1

answer and explanation

we are given:

V1= 15 l

T1 = 51 deg

P1 = ?

P2= 3.9 atm

V2= 25 litres

T2 = 41 deg

we can use the combine gas equation to find initial

P1V1/T1 = P2V2/T2

we can the rearrange the equation to make P1 the subject of the equation

P1 = P2V2T1/V1T2

= 3.9 x 25 x 51 / 15 x41

= 8.1 atm

the initial pressure of the gas was 8.1 atm