Combined Gas Law You can re-write the ideal gas law P 1 V 1 = P 2 V 2 n 1 T 1 n 2 T 2

Example – A nitrogen cylinder contains 3.42 kg of nitrogen at psi absolute and 20.0 o C. What is the pressure if the temperature is 150. o C, but you have released 0.20 kg of nitrogen? 2718 psi ≈ 2720 psi

Example – A nitrogen cylinder contains 3.42 kg of nitrogen at psi absoluteand 20.0 o C. What is the pressure if the temperature is 150. o C, but you have released 0.20 kg of nitrogen? n1 = 3.42 kg, n2 = = 3.22 kg (since it is proportions we don’t need to calculate mols!! T1 = = K, T2 = = K P1 = psi, P2 = ???? PV = PV nt nT (2000 psi)(V) = P2V (V cancels) (3.42)(293.15) (3.22)(423.15) P2 = 2718 psi ≈ 2720 psi

PV = nRT (1.00)V = nR(283) (1.15)V = nR(T??) 52.5 o C An airtight drum at 1.00 atm and 10.0 o C is heated until it reaches a pressure of 1.15 atm. What is the new temperature in o C? (3)

PV = nRT (162)(14.5) = nRT ( P )(17.2) = nRT 137 Jukkalas An airtight cylinder has a pressure of 162 Jukkalas when the piston is 14.5 cm from the bottom. What is the pressure if the piston is moved to 17.2 cm from the bottom of the cylinder? (Assume that the temperature is the same)

P = P gauge + 1 atm (1 atm = kPa) p 1 = p 2 T 1 T 2 T 1 = = 283 P 1 = = kPa = x 10 3 Pa T 2 = = 325 K P 2 = ??? 211 kPa Absolute, 109 kPa Gauge A tyre is at 82 kPa gauge pressure when the temperature is 10.0 o C. What is the gauge pressure if the temperature is 52 o C (assume the volume remains constant, and that the tyre does not leak)