1. Ideal Gas LawPV = nRT Application of PV=nRT Density Molecular Mass

4. T(K) = 273.15+ t(0C) 1atm= 14.7 psi

5. R= 0.08206 (atm.L)/(mol.K) 2 1 Knowing PV= nRT, Rearrange the equation to : 3

6. Standard Conditions The Standard Temperature & pressure, STP is when Standard pressure = 1 atm Standard temperature = 273 K, 0 °C 1 mole of any gas at STP condition has a volume equal to 22.4 L

7. Molar Volume = Volume of 1 mole of a gas Solving the ideal gas equation for the volume of 1 mole of any gas at STP gives a volume of 22.4 L 1 mole of a gas has 6.022x1023 molecules of gas 6.022x1023 molecules of gas occupies a volume of 22.4L at STP Molar volume of a gas at STP is 22.4 L One mole of different gases have the same volume at same temperature and pressure, but have different masses

8. Molar Volume and Molar mass 4.0 g 131.23 g 16.05 g 6.022x1023atomsHe6.022x1023atomsXe 6.022x1023 moleculesCH4 Different molar mass, same molar volume at same T &P

9. Example 2: A gas occupies 10.0 L at 44.1 psi and 27 °C. Calculate the volume this gas occupies at standard conditions using Ideal Gas law V1 =10.0L, P1= 44.1psi, t1= 27°C, P2 =1.00atm, t2=0°C and V2=?L T(K) = 273.15+ t(0C) 1atm= 14.7 psi PV= nRT and R= 0.08206 (atm.L)/(mol.K)

10. Example 2: Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C mSO2 = 637 g, P = 6.08 x 104 mmHg, t= −23 °C, &V=?L 1 atm= 760 mmHg 1 mole of SO2 = 64.07 g SO2 T(K) = 273.15+ t(0C) PV= nRT & R= 0.08206 (atm.L)/(mol.K)

11. Density at Standard Conditions Density is the ratio of mass to volume Density of a gas is generally given in g/L The mass of 1 mole = molar mass The volume of 1 mole at STP = 22.4 L Example: Calculate the density of N2(g) at STP

15. What is the molar mass of a gas if 12.0 g occupies 197 L at 3.80 x 102 torr and 127 °C? m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K, molar mass= ?g/mol 1atm =760 mmHg, T(K)=273.15+t(0C), PV=nRT, & Molar Mass =(mass/n) g/mole