Gas Laws and Partial Pressures

Gas Laws: Exercise 1

  • Given: p=101.3kPap = 101.3 kPa, V=10LV = 10 L, T=25°CT = 25 °C, n=??n = ??
  • Ideal Gas Equation: pV=nRTpV = nRT
  • R=8.314LkPaK−1mol−1R = 8.314 L kPa K^{-1} mol^{-1}
  • Convert Celsius to Kelvin: T<em>Kelvin=T</em>celsius+273T<em>{Kelvin} = T</em>{celsius} + 273
  • T=25+273K=298KT = 25 + 273 K = 298 K
  • Calculation: n = rac{pV}{RT} = rac{101.3 kPa imes 10 L}{8.314 J K^{-1} mol^{-1} imes 298 K} = 0.409 mol

General Gas Equation: Exercise 2

  • Given: p<em>1=100kPap<em>1 = 100 kPa, p</em>2=250kPap</em>2 = 250 kPa, T<em>1=25°C=298KT<em>1 = 25 °C = 298 K, T</em>2=?T</em>2 = ? (V and n are constant)
  • General Gas Equation: rac{p1}{T1} = rac{p2}{T2}
  • T_2 = rac{298 K imes 250 kPa}{100 kPa} = 745 K (472°C)

Ideal Gas Equation: Exercise 3

  • Given: p=200kPap = 200 kPa, T=120°C=393KT = 120 °C = 393 K, V=5.0LV = 5.0 L, n=??n = ??
  • Ideal Gas Equation: pV=nRTpV = nRT
  • n = rac{pV}{RT} = rac{200 kPa imes 5.0 L}{8.314 L kPa K^{-1} mol^{-1} imes 393 K} = 0.306 moles
  • Converting moles to mass: m=nM=0.306molesimes18gmol−1=5.5gm = nM = 0.306 moles imes 18 g mol^{-1} = 5.5 g

Dalton's Law of Partial Pressures: Exercise 4

  • Given: n<em>O</em>2=2n<em>{O</em>2} = 2, n<em>N</em>2=8n<em>{N</em>2} = 8, Total # moles = 10, Ptotal=100kPaP_{total} = 100 kPa
  • Dalton's Law: P<em>A=x</em>AP<em>totalP<em>A = x</em>A P<em>{total}, where xA = rac{nA}{n{total}}
  • x{O2} = rac{2}{8+2} = 0.2, P<em>O</em>2=0.2imes100kPa=20kPaP<em>{O</em>2} = 0.2 imes 100 kPa = 20 kPa
  • x{N2} = rac{8}{8+2} = 0.8, P<em>N</em>2=0.8imes100kPa=80kPaP<em>{N</em>2} = 0.8 imes 100 kPa = 80 kPa