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Unit 6: Gases

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Unit 6: Gases

Section 2: Ideal Gas Law

- Stands for standard temperature and pressure
- Standard temperature is 0ºC or 273 K
- Standard pressure is 1 atm or 760 mm or 760 torr

- Necessary for scientists to have some type of standard so they can compare results

- Suppose you perform an experiment in your lab and have 500 mL of oxygen at 25ºC and 750 mm of pressure. What is the volume if the experiment is conducted at STP?
- P1V1 = P2V2
T1 T2

- (750 mm)(500 mL) = (760 mm) V2(P2 and T2 are from STP)
25 + 273 273

- 452 mL = V2

- P1V1 = P2V2

- Also known as “PIV NERT” because of its formula: PV=nRT
- P = pressure
- V = volume (in liters)
- n = number of moles of the gas
- R = a constant whose value is 0.0821 if the pressure is in atm, or 62.3 if the pressure is in mm or torr
- T = temperature (in kelvins)

- What pressure (in millimeters), is expected from 0.55 mole of nitrogen that is at 20ºC and in a 0.75 L container?
- PV = nRT
- R = 62.3 mm since we are asked to find pressure in millimeters

- (P)(0.75 L) = (0.55 mol)(62.3 mm)(20+273 K)
- 0.75 P = 10039.645
- P = 13,386.19 mm

- PV = nRT

- Ideal gas law can be used to find molecular weight
- Molecular weight = grams
mole

- Molecular weight = grams
- When you know the number of grams and can use the ideal gas law to calculate moles, you can find molecular weight

- Calculate the molecular weight of an unknown gas if 0.35 g of it occupies a volume of 0.1 L at a pressure of 700 mm and a temperature of 298 K.
- Step 1: Calculate moles
- PV = nRT
- (700 mm)(0.1 L) = (n)(62.3)(298 K)
- 70 = 18565.4 n
- 0.0038 mole = n

- Step 2: Calculate molecular weight using gram/mole
- Molecular weight = 0.35 g = 92 grams/mole
0.0038 mol

- Molecular weight = 0.35 g = 92 grams/mole