The pressure gauge on Tank #3 reads 2.50 atm. What is the identity of the monatomic gas in Tank #3? The temperature is 25C, and the Volume is 50L, and Tank 3 has 671 grams of unknown substance.
This is the question, it is not relevant to another part of the question, it is it's own question. I am just confused on how to find the answer. Can someone please help me set up this problem? Thanks!
This is the question, it is not relevant to another part of the question, it is it's own question. I am just confused on how to find the answer. Can someone please help me set up this problem? Thanks!
-
First, use PV=nRT to calculate the number of moles, n, in the tank.
n = PV/RT
n = (2.50 atm)(50 L) / ( 0.082057 L atm/mol K)(298 K)
n = 5.11 mol
Second, use the given mass and n to calculate the molar mass of the gas.
671 g / 5.11 mol = 131 g/mol
Third, determine which element matches this molar mass.
Since 131 is very close to the atomic mass of Xe, Xe is the unknown gas.
n = PV/RT
n = (2.50 atm)(50 L) / ( 0.082057 L atm/mol K)(298 K)
n = 5.11 mol
Second, use the given mass and n to calculate the molar mass of the gas.
671 g / 5.11 mol = 131 g/mol
Third, determine which element matches this molar mass.
Since 131 is very close to the atomic mass of Xe, Xe is the unknown gas.