If I contain 3 moles of gas in a container with a volume of 60 liters and at a

temperature of 400 K, what is the pressure inside the container? (Use mm Hg)

P = 1246.4 mmHg

Explanation:

assuming ideal gas:

PV = RTn

∴ R = 0.082 atm. L/K. mol

∴ V = 60 L

∴ T = 400 K

∴ n gas = 3 mol

⇒ P = RTn / V

⇒ P = ((0.082 atm. L/K. mol) (400 K) (3 mol)) / (60 L)

⇒ P = (1.64 atm) * (760 mmHg/atm)

⇒ P = 1246.4 mmHg

Answer: n = 3.0 moles

V = 60.0 L

T = 400 K

From PV = nRT, you can find P

P = nRT/V = (3.0 mol) (0.0821 L-atm/K-mol) (400 K) / 60.0L

P = 1.642 atm = 1.6 atm (to 2 significant figures)