a 4.40 l sample of neon at 7.10 atm is added to a 13.0 l cylinder that contains argon. if the pressure in the cylinder is 7.35 atm after the neon is added, what was the original pressure (in atm) of argon in the cylinder?



Answer :

The original pressure (in atm) of argon in the cylinder is 8.062atm

Volume of neon = 4.40L

Volume of argon = 13L

total volume = 17.4L

partial pressure of neon = 4.40l * 4.22atm/17.4L = 1.1552 atm [ partial pressure = original pressure X (volume of the gas/total volume)]

total pressure = 7.01

partial pressure of argon = 5.8548 atm

original pressure of argon = 5.8548*17.9/13 = 8.062atm

Each gas that makes up a mixture of gases has a partial pressure, which is the notional pressure of that gas as if it alone filled the original combination's complete volume at the same temperature.

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