Question

A cylinder of 20.0L capacity contains 160g of oxygen gas at 25degree C what mass of oxygen should be released to reduce the pressure of cylinder to 1.2atm​

Answer 1

Explanation:

Moles of gas that will be present at 1.2 atm will be

Moles of gas that will be present at 1.2 atm will ben=PV/RT=(1.2 ×20)/(0.0821×298)=0.98 mol

Moles of gas that will be present at 1.2 atm will ben=PV/RT=(1.2 ×20)/(0.0821×298)=0.98 molmass of oxygen gas = 0.98 mol×32

Moles of gas that will be present at 1.2 atm will ben=PV/RT=(1.2 ×20)/(0.0821×298)=0.98 molmass of oxygen gas = 0.98 mol×32 gmol^−1=31.36 g

Moles of gas that will be present at 1.2 atm will ben=PV/RT=(1.2 ×20)/(0.0821×298)=0.98 molmass of oxygen gas = 0.98 mol×32 gmol^−1=31.36 gTherefore, the mass of oxygen gas that must be removed = 160 - 31.36 = 128.64 g