Question

A gas that has a volume of 28 Liters, a Temperature of 318 K, and an unknown pressure initially

has its Volume increased to 34 Liters and its Temperature decreased to 308 K. If I measure the
pressure after the change to be 20 atm, what was the original pressure of the gas?

Answer 1

Answer: 25.07 atm

Step-by-step explanation:

Given that,

Initial Pressure of gas (P1) = ?

Initial Volume of gas (V1) = 28L

Initial Temperature of gas (T1) = 318K

Final pressure of helium = 20 atm

Final volume of gas (V2) = 34L

Final temperature of gas (T2) = 308K

Since pressure, volume and temperature are given, apply the combined gas equation

(P1V1)/T1 = (P2V2)/T2

(P1 x 28L)/318K = (20 atm x 34L)/308K

28L•P1/318K = 680 atm•L/308K

To get P1, cross multiply

28L•P1 x 308K = 680 atm•L x 318K

8624L•K•P1 = 216240 atm•L•K

Divide both sides by 8624L•K

8624L•K•P1/8624L•K = 216240 atm•L•K/8624L•K

P1 = 25.07 atm

Thus, the original pressure of the gas is 25.07 atmosphere