Review for Test 5

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# Review for Test 5 - PowerPoint PPT Presentation

Review for Test 5. Gas Laws. Charles Law V 1 = V 2 P constant T 1 T 2 Boyles Law P 1 V 1 = P 2 V 2 T constant Gay-Lussac’s Law P 1 = P 2 V constant T 1 T 2 Combined P 1 V 1 = P 2 V 2

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## Review for Test 5

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### Review for Test 5

Gas Laws

Charles LawV1 =V2 P constant

T1 T2

Boyles Law P1V1 = P2V2 T constant

Gay-Lussac’s Law P1=P2 V constant

T1 T2

CombinedP1V1=P2V2

T1 T2

Ideal Gas Law PV = nRT

### Dalton’s Law of Partial Pressure

The sum of the individual gas pressures equals the overall pressure of the mixture of gases.

P1 + P2 + P3 . . . = Ptotal

A sample of gas occupies 2.97 L at 72˚C.

What volume would it take up at 502K?

V1 =V2

T1 T2

Volume should increase!

2.97L x 502K = 4.32 L

345K

A gas-filled bottle is heated from 25°C to 89°C. If the original pressure was 760 torr, what would be the new pressure ?

T1 = 25+273=298K

T2 = 89+273=362K

760 = P2

298 362

P2 = 923 torr

A gas at 98.2 kPa has a volume of 39L. What is its volume at 120.5kPa and constant temperature?

P1V1 = P2V2

Volume should decrease.

39L x 98.2 kPa = 31.8L

120.5 kPa

At STP, the volume of a gas is 325 mL. What volume does it occupy at 20.0°C and 93.3kPa?

P1V1=P2V2

T1 T2

325 mL x 293K x 101.3 kPa = 379 mL

273K 93.3 kPa

What pressure is exerted by .00306 mole of gas in a 25.9 cm3 container at 9˚C?

P= nRT

V

= .00306mol x 8.31 L∙kPa x 282K

.0259 L mol∙K

= 277kPa

A 35.0L tank contains 7.00 mol of compressed air. If the pressure inside the tank is 500.0 kPa, what is the temperature of the compressed gas?

T = PV

nR

T = 500.0kPa x 35.0L

7.00mol x 8.31

= 301 K

A gaseous mixture containing argon, nitrogen, and oxygen is in a vessel. If the total pressure is 98.5 kPa, and the partial pressures of nitrogen and argon are 22.0 kPa and 50.0 kPa respectively, what is the partial pressure of oxygen?

Ptot = PAr + PN + PO

98.5 kPa = 22.0 + 50.0 + Po

PO = 26.5 kPa