Question

A sealed jar has 0.20 moles of gas at a pressure of 300.12 kpa and a temperature of 229 k. what is the volume of the jar?

Answer 1

From ideal gas equation v=nRT/P where n is number of moles, ,R is the gas constant, T is the temperature and P is the pressure
P=300.12kpa
n=0.2moles
R(the gas constant)=8.3145 L.kpa/mol.k
T=229k
v=?
0.2mol x { ( 8.1345 l.kpa/mol.k x229 k)/300.12 kpa}=1.27 L

Answer 2

We are going to consider that is a ideal gas and then it accomplish the following equation:
P*V = n*R*T
where P is pressure, V is volume, n is mols, T is temperature and R is a constant value of 0.082 (atm*l)/(mol*K)
First we need to change the units of pressure 300.12 kpa = 2.96 atm
Then:
V = 0.2 * 0.082 * 229 / 2.96 = 1.27 liters
So the volume of the jar is 1.27 liters