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