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?

25.07 atm

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

the atomic number will stay the same no matter how many electrons it has, so it doesn’t decrease nor increase. it’ll only change with a different number of protons