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The Ideal Gas Law

The Ideal Gas Law. PV = n RT. Definition. The ideal gas law is the mathematical relationship among: Pressure (P) Volume (V) Temperature (T) Amount of the Gas in Moles (n) The law quantitatively describes the behavior of a gas sample for any combination of conditions:

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The Ideal Gas Law

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  1. The Ideal Gas Law

    PV = nRT
  2. Definition The ideal gas law is the mathematical relationship among: Pressure (P) Volume (V) Temperature (T) Amount of the Gas in Moles (n) The law quantitatively describes the behavior of a gas sample for any combination of conditions: If you know three of the conditions, you can solve for the fourth. PV = nRT
  3. Units of Measurement Volume must always be measured in liters (L). 1L = 1000mL = 0.001kL Amount of the gas must always be measured in moles (mol). If amount is given in grams, you must do a gram to mole conversion! Temperature must always be measured in Kelvin (K). K = ˚C + 273 ˚C = K - 273 Pressure must always be measured in atmospheres (atm). 1 atm= 760 mmHg = 760 torr= 1.013 x 105 Pa = 101.3 kPa
  4. Standard Temperature and Pressure (STP) Standard temperature is 273 K. Standard pressure is 1 atm. At STP, 1 mole of any ideal gas occupies 22.4 L.
  5. The Ideal Gas Constant Represented as R in the ideal gas law. R is a constant value that never changes. The value for R is derived from the standard volume (22.4L) of 1 mole of a gas at STP. PV = nRT can be arranged as R = PV/nT R = (1 atm)(22.4 L)/(1 mol)(273 K) = 0.0821 L•atm/mol•K PV = nRT
  6. Solving Ideal Gas Law Problems What is the pressure exerted by a 0.500 mol sample of nitrogen gas in a 10.0L container at 25˚C? 1. Determine given values and unknown. n= 0.500 mol, V = 10.0 L, T = 25˚C , R = 0.0821 L•atm/mol•K P = ? 2. Convert given values into proper units. K = 25˚C + 273 = 298 K 3. Rearrange equation to solve for unknown. Plug in given values to solve for unknown. P = nRT/V P = (o.500 mol)(0.0821 L•atm/mol•K)(298 K) / 10.0 L P = 1.22 atm
  7. Solving Ideal Gas Law Problems What is the volume of an 8.00g sample of oxygen gas at 293 K and 0.974 atm? 1. Determine given values and unknown. n= 8.00g, T = 293 K , P = 0.974 atm, R = 0.0821 L•atm/mol•K V= ? 2. Convert given values into proper units. 8.oo g O2 x (1 mol O2 / 32.00 g O2 ) = 0.250 mol O2 3. Rearrange equation to solve for unknown. Plug in given values to solve for unknown. V = nRT/P V= (o.250 mol)(0.0821 L•atm/mol•K)(293 K) / 0.974 atm V = 6.17 L
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