Gases Properties PDF

Chemical Properties of Gases  Molar Volume Which measurement of a gas is best to use – MASS or VOLUME? Why? Volume Easier Generally larger More precise Avogadro's Hypothesis A B Same V Same T & P Motion is same Contain same # of particles...

Chemical Properties of Gases  Molar Volume Which measurement of a gas is best to use – MASS or VOLUME? Why? Volume Easier Generally larger More precise Avogadro's Hypothesis A B Same V Same T & P Motion is same Contain same # of particles A specific number of particles in 2 gases under same T & P will occupy same V What # of particles is convenient to use? Mole 6.02 x 1023 particles 1 mole Molar Volume (Vm) V of 1 mole of ANY gas @ specific T & P @ STP (O°C, 101kPA): 1 mole gas = 6.02 x 1023 molecules = 22.4L 1 22.4L mole 1 22.4 L mole @ SATP (25°C, 100kPA): 1 mole gas = 6.02 x 1023 molecules = 24.8 L 1 24.8L mole 1 Molar volume allows conversion between volume (V) and moles (n) And mass (m) Molar V molar mass V  n  m molar Molar V mass m  n  V Calculate the volume of 0.024 mol of carbon dioxide gas at SATP. 0.024 molx 24.8 L = 0.60 L 1 mole Calculate the chemical amount (moles) of 5.6L of O2 at STP. 5.6 Lx 1 mole = 0.25 mole 22.4 L Mass? 0.25 molex 32.00 g= 8.0 g 1 Practice Molar Volume #1-6  Combining Volumes In a reaction, the VOLUME of gaseous reactants & products are always in simple ratios of whole #s T & P are constant H2 + Cl22(g) → HCl (g) (g) Ratio: 1 : 1 : 2 1L 1L 2L 100mL 100mL 200mL Not all reactants/products need to be gases But the law ONLY applies to gases 2H2O (l) → 2H2 (g) + O2 (g) Ratio: 2 2 1 2L 1L O2 H2 What VOLUME of oxygen would need to completely combust 120ml of butane (C4H10) gas from a lighter? C4H10 (g) + O2 (g) → CO2 (g) + H2O (g) 2C4H10 (g) + 13O2 (g) → 8CO2 (g) + 10H2O (g) 120ml V? O2 reacts with C4H10 in a 13:2 ratio 13 L of O2 reacts with 2 L of C4H10 13 ml of O2 x reacts with 13 ml O2 2=ml780ml of C4HO10 2 120 ml C H 2 ml C4H10 4 10 What VOLUME of CO2 (g) is produced when 120ml of butane (C4H10) gas from a lighter is completely combusted? 2C4H10 (g) + 13O2 (g) → 8CO2 (g) + 10H2O (g) 120ml V? 2:8 ratio 2 L of C4H10 produces 8 L of CO2 2 ml of C4H10 produces 8 ml of C4H10 x 8 ml CO2 = 480 ml CO2 120 ml C4H10 2 ml C4H10 Practice #1-5 Ideal Gas Law  Ideal Gas Real Gas Hypothetical Molecules are very far Molecules are forced apart close together Size of molecules are significant Molecules experience Molecules in constant, intermolecular forces random, straight-line As T , molecules slow motion down No forces between them Collisions are not perfectly elastic Molecules undergo Energy is lost perfect elastic collisions Pressure is less No energy is lost Real gases can behave as ideal gases Low P & high T STP / SATP Can describe inter-relationship of P, T, V & n using a single equation PV=nRT P in kPa V in L n = moles T in K R = universal gas constant *Determine R by using STP conditions for 1 mole of ideal gas PV = nRT R = PV @ STP 0°C nT 1 mole = 101kPa 22.4L R = 101.325 kPa (22.414 L) 1 mol (273.15 k) R = 8.314 kPa  L mol  K Often mass is the starting point… Covert mass to moles Use PV = nRT Or ending point… Use PV = nRT Covert moles to mass What is the volume of 0.78g of hydrogen gas @ 22°C & 125kPa? 0.78gx 1 mole =0.39 mol 2.02g PV = nRT V = nRT P V = 0.39mol (8.314kPaL/molK) (295K) 125kPa V = 7.6L What mass of argon gas is needed in a 0.88L tube to produce SATP conditions? PV = nRT n = PV RT n = 100kPa (0.88L) 8.314kPaL/molK (298K) n = 0.035519 mol 0.035519mol x 39.95g m = 1.4g mol Practice Ideal Gas Law #1-6 #7 – Calculating molar mass Hint: useful practice for lab… #8 - Additional practice