Question

help pleasee

How many grams of Kr are in a 6.88 L cylinder at 62.5 'C and 3.12 atm? mass:

Answer 1

using the ideal gas equation

PV =nRT

n = mass / molar mass

PV = mass * R T / molar mass

=> mass = PV * molarmass / RT

P = 3.12 atm

V = 6.88 L

R = 0.0821 L atm / k mol

T = 62.5 c = 335.5 K

molar mass = 83.798 g/mol

mass = 3.12 * 6.88 * 83.798 / 0.0821 * 335.5

mass => 65.3 grams

=> no of moles = mass / molar mass

for Ar = 4.05g / 39.948 g/mol => 0.1014 moles

V = 2.00 L

R = 0.0821 l atm /k mol

T = 27c = 300 k

using the ideal gas equation PV = nRT

P = nRT / V

P = 0.1014 * 0.0821 * 300 / 2

Partial pressure of Ar = 1.2485 atm

partial pressure of He = 2.85 atm

total pressure = Partial pressure of Ar + partial pressure of He => 1.2485 + 2.85 => 4.098 atm

answer => 4.1 atm