Learning about the special behavior of gases The Gas Laws Learning about the special behavior of gases Objective #3 The Ideal Gas Law, pg. 6

The Ideal Gas Law This equation considers a fourth variable, the NUMBER of particles, and the “ideal gas constant” (a.k.a. Avogadro’s Hypothesis) into the combined gas law. It would be good to familiarize yourself with the content poster.

The Ideal Gas Law PV = nRT This equation considers a fourth variable… the NUMBER of particles! This equation incorporates Avogadro’s Hypothesis into the combined gas law. PV = nRT

Solving for “R” Use the values of standard condition A: Using Pressure in kPa

Solving for “R” A: Using Pressure in kPa R = Pressure(P) x Volume(V) # of mol(n)xTemp(K)

Solving for “R” A: Using Pressure in kPa R = Pressure(P) x Volume(V) = (101.3kPa)x(22.4L) # of mol(n)xTemp(K) = (1 mol)x(273K)

Solving for “R” A: Using Pressure in kPa R = Pressure(P) x Volume(V) = (101.3kPa)x(22.4L) # of mol(n)xTemp(K) = (1 mol)x(273K) R = 8.31 kPa  L Mol  K

Solving for “R” B: Using Pressure in ATM

Solving for “R” B: Using Pressure in ATM R = Pressure(P) x Volume(V) # of mol(n)xTemp(K)

Solving for “R” B: Using Pressure in ATM R = Pressure(P) x Volume(V) = (1ATM)x(22.4L) # of mol(n)xTemp(K) = (1 mol)x(273K)

Solving for “R” B: Using Pressure in ATM R = Pressure(P) x Volume(V) = (1ATM)x(22.4L) # of mol(n)xTemp(K) = (1 mol)x(273K) R = 0.0821 ATM  L Mol  K

Solving for “R” C: Using Pressure in mm Hg R = (P) x (V) (n)x(K)

Solving for “R” C: Using Pressure in mm Hg R = (P) x (V) = (760mmHg)x(22.4L) (n)x(K) = (1 mol)x(273K)

Solving for “R” C: Using Pressure in mm Hg R = (P) x (V) = (760mmHg)x(22.4L) (n) x (K) = (1 mol)x(273K) R = 62.4 mmHg  L Mol  K

Example 1, pg. 7 You fill a rigid steel cylinder that has a volume of 20 L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain? How many grams is this?

Example 1 You fill a rigid steel cylinder that has a volume of 20 L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain? Rearrange the formula to find what we’re looking for: How many grams is this?

Example 1 You fill a rigid steel cylinder that has a volume of 20 L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain? Rearrange the formula to find what we’re looking for: n = PV RT How many grams is this?

Example 1 You fill a rigid steel cylinder that has a volume of 20 L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain? Rearrange the formula to find what we’re looking for: n = PV = (20,000kPa)x(20L) = RT (8.31 kPaL/molK)x(301K) How many grams is this?

Example 1 You fill a rigid steel cylinder that has a volume of 20 L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain? Rearrange the formula to find what we’re looking for: n = PV = (20,000kPa)x(20L) = _________ mol N2 RT (8.31 kPaL/molK)x(301K) How many grams is this? / / / / / /

Example 1 You fill a rigid steel cylinder that has a volume of 20 L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain? Rearrange the formula to find what we’re looking for: n = PV = (20,000kPa)x(20L) = 159.92 mol N2 RT (8.31 kPaL/molK)x(301K) How many grams is this? / / / / / /

Example 1 You fill a rigid steel cylinder that has a volume of 20 L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain? Rearrange the formula to find what we’re looking for: n = PV = (20,000kPa)x(20L) = 159.92 mol N2 RT (8.31 kPaL/molK)x(301K) How many grams is this? 159.92 mol N2 28 g N2 = 1 1 mol N2 / / / / / /

Example 1 You fill a rigid steel cylinder that has a volume of 20 L with nitrogen gas to a final pressure of 20,000 kPa at 28o C. How many moles of nitrogen gas does the cylinder contain? Rearrange the formula to find what we’re looking for: n = PV = (20,000kPa)x(20L) = 159.92 mol N2 RT (8.31 kPaL/molK)x(301K) How many grams is this? 159.92 mol N2 28 g N2 = 4,477.7 g N2 1 1 mol N2 / / / / / /

Example 2 A deep underground cavern contains 2.24 x 106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain?

Example 2 A deep underground cavern contains 2.24 x 106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain? n = PV RT

Example 2 A deep underground cavern contains 2.24 x 106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain? n = PV = (1,500kPa)x(2.24 x 106L) RT (8.31 kPaL/molK)x(315K)

Example 2 A deep underground cavern contains 2.24 x 106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain? n = PV = (1,500kPa)x(2.24 x 106L) = 1.28X106 mol CH4 RT (8.31 kPaL/molK)x(315K)

Example 2 A deep underground cavern contains 2.24 x 106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain? n = PV = (1,500kPa)x(2.24 x 106L) = 1.28X106 mol CH4 RT (8.31 kPaL/molK)x(315K) 1.28X106 mol CH4 x 16g CH4 1 1 mol CH4

Example 2 A deep underground cavern contains 2.24 x 106 L of methane gas, CH4, at a pressure of 1.5 x 103 kPa and a temperature of 42o C. How many grams of methane does this natural-gas deposit contain? n = PV = (1,500kPa)x(2.24 x 106L) = 1.28X106 mol CH4 RT (8.31 kPaL/molK)x(315K) 1.28X106 mol CH4 x 16g CH4 = 2.05X107 g CH4 1 1 mol CH4

Example 3. When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 10^3 kPa. How many moles of helium does the sphere contain?

Example 3. When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 10^3 kPa. How many moles of helium does the sphere contain? n = PV RT

Example 3. When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 10^3 kPa. How many moles of helium does the sphere contain? n = PV = (1.89x103 kPa)x(685L) RT (8.31 kPaxL/molxK)x(621K)

Example 3. When the temperature of a rigid hollow sphere containing 685 L of helium gas is held at 621 K, the pressure of the gas is 1.89 x 10^3 kPa. How many moles of helium does the sphere contain? n = PV = (1.89x103 kPa)x(685L) = 250.88 moles He RT (8.31 kPaxL/molxK)x(621K)

Example 4 What pressure will be exerted by 0.45 mol of a gas at 25o C if it is contained in a 0.65 L vessel?

Example 4 What pressure will be exerted by 0.45 mol of a gas at 25o C if it is contained in a 0.65 L vessel? P = nRT V

Example 4 What pressure will be exerted by 0.45 mol of a gas at 25o C if it is contained in a 0.65 L vessel? P = nRT = (0.45 mol)x(8.31 kPaL/ molK)x(298K) V 0.65 L

Example 4 What pressure will be exerted by 0.45 mol of a gas at 25o C if it is contained in a 0.65 L vessel? P = nRT = (0.45 mol)x(8.31 kPaL/molK)x(298K) =1714.4 kPa V 0.65 L

Example 5 A child has a lung capacity of 2.2 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole.

Example 5 A child has a lung capacity of 2.2 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole. n = PV RT

Example 5 n = PV = __(102kPa)x(2.2L)__ RT (8.31kPaL/molK)x(310K) A child has a lung capacity of 2.2 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole. n = PV = __(102kPa)x(2.2L)__ RT (8.31kPaL/molK)x(310K)

Example 5 n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air A child has a lung capacity of 2.2 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole. n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air RT (8.31kPaL/molK)x(310K)

Example 5 n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air A child has a lung capacity of 2.2 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole. n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air RT (8.31kPaL/molK)x(310K) 0.87 mol air x 28g air 1 1 mol air

Example 5 n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air A child has a lung capacity of 2.2 L. How many grams of air do her lungs hold at a pressure of 102 kPa and a normal body temperature of 37o C? Air is a mixture, but you may assume it has a molar mass of 28 g/mole. n = PV = __(102kPa)x(2.2L)__ = 0.087 mol air RT (8.31kPaL/molK)x(310K) 0.87 mol air x 28g air = 2.4g air 1 1 mol air

Example 6 What volume will 12 grams of oxygen gas occupy at 25o C and a pressure of 52.7 kPa?

Example 6 12g O2 x 1 mol O2 = 0.375 mol O2 1 32g O2 17.62 L O2 What volume will 12 grams of oxygen gas occupy at 25o C and a pressure of 52.7 kPa? 12g O2 x 1 mol O2 = 0.375 mol O2 1 32g O2 V = nRT = (0.375 molO2)x(8.31kPaL/molK)x(298K) = P (52.7 kPa) 17.62 L O2

It’s always a good idea to regularly review the notes we’ve take up to this point.