The Ideal Gas Law

QuickNotes

Ideal Gas Law: PV = nRT

• P = pressure (atm)
• V = volume  (L)
• n = moles of gas (mol)
• R = universal gas constant (0.08206 L x atm/K x mol)
• T = temperature (K)
• Important Concept: The units of each component of the equation MUST be in the units given above so that they correspond with the universal gas constant. 
• This law essentially combines all of the basic gas laws into one coherent formula
• It is known as the equation of state for a gas at a given condition at a given time
• There is no such thing as a real ideal gas. Gases that behave nearly ideally do so at low pressures and high temperatures. 

Examples: 

Example 1: Determine the volume of occupied by 2.34 grams of carbon dioxide gas at STP.

• Rearrange PV = nRT to this: V = nRT / P
• Substitute:

V = [ (2.34 g / 44.0 g mol¯¹) (0.08206 L atm mol¯¹ K¯¹) (273.0 K) ] / 1.00 atm

 V = 1.19 L (to three significant figures)

Example 2: A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles of argon and the mass in the sample.

• Rearrange PV = nRT to this: n = PV / RT
• Substitute:

n = [ (1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯¹ K¯¹) (273.0 K) ]

 n = 2.50866 mol (I'll keep a few guard digits)

• Multiply the moles by the atomic weight of Ar to get the grams:

2.50866 mol x 39.948 g/mol = 100. g (to three sig figs)

Example 3: 96.0 g. of a gas occupies 48.0 L at 700.0 mm Hg and 20.0 °C. What is its molecular weight?

• Solve for the moles using PV = nRT: n = PV / RT

 n = [ (700.0 mmHg / 760.0 mmHg atm¯¹) (48.0 L) ] / [ (0.08206 L atm mol¯¹ K¯¹) (293.0 K) ]

 n = 1.8388 mol

• Divide the grams given (96.0) by the moles just calculated above:

96.0 g / 1.8388 mol = 52.2 g/mol

Watch this review video to completely understand the ideal gas law!

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Ideal Gas Law Quiz