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Ideal Gas Law PV = nRT

Ideal Gas Law PV = nRT. Application of PV= nRT Density Molecular Mass. Ideal Gas Law. By combining the gas laws we can write a general equation where P is the pressure , V is the volume, n is the amount of gas in mole, and T(K ) is the temperature

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Ideal Gas Law PV = nRT

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  1. Ideal Gas LawPV = nRT

    Application of PV=nRT Density Molecular Mass
  2. Ideal Gas Law By combining the gas laws we can write a general equation where P is the pressure , V is the volume, n is the amount of gas in mole, and T(K) is the temperature The constant is shown byR, gas Law constant . The value of R, depends on the units of P & V R =0.08206[(atm.L)/(mol.K)], when P is in atm and V is in L
  3. Example: How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C? V = 3.24 L, P = 24.3 psi, t = 25 °C and n=? Mol T(K) = 273.15+ t(0C) 1atm= 14.7 psi R= 0.08206 (atm.L)/(mol.K) 2 1 Knowing PV= nRT, Rearrange the equation to : 3
  4. Standard Conditions 1 mole of any gas at STP condition has a volume equal to 22.4 L The Standard Temperature & pressure, STP is when Standard pressure = 1 atm Standard temperature = 273 K, 0 °C
  5. 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
  6. 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
  7. 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)
  8. 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)
  9. Density at Standard Conditions Example: Calculate the density of N2(g) at STP 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
  10. Gas Density Density is directly proportional to molar mass As Molar mass of a gas increases, its density is also increase when T, n, P, & V are constant
  11. Example : Calculate the density of N2 at 125°C and 755 mmHg P = 755 mmHg, t = 125 °C, dN2 = ?g/L 1 atm =760 mmHg, 1mole N2=28.0g N2, T(k) = 273.15 + t(0C)
  12. Example2: Calculate the density of a gas at 775 torr and 27 °C if 0.250 moles weighs 9.988 g m = 9.988g, n = 0.250 mol, P = 1.0197 atm, T = 300.K Density= ?g/L 1atm =760 mmHg, T(K)=273.15+t(0C), PV=nRT, & d=(m/v)
  13. Molar Mass of a Gas One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law to calculate the number of moles, then
  14. Example: Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg m=0.311g, V=0.225L, P=1.1658 atm, T=328 K, molar mass= ? g/mol 1atm =760 mmHg, T(K)=273.15+t(0C), PV=nRT, & Molar Mass =(mass/n) g/mole
  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
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