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Published byChester Paul Modified over 8 years ago
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The Combined Gas Law
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The relationship among pressure, volume, and temperature can be mathematically represented by an equation known as the combined gas law. P 1 V 1 = P 2 V 2 T 1 T 2 where: P 1 is the initial pressure and P 2 is the new pressure. V 1 is the initial volume and V 2 is the new volume. T 1 is the initial temperature and T 2 is the new temperature. Temperature must always be in Kelvin!
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Ex. (1) If 22.4 L of a gas at 0.0 o C and 1.29 atm were cooled to a temperature of -30.0 o C as it was expanded to a volume of 85.9 L, what would be its new pressure in atm? (X)(22.4 L) 273.0 K243.0 K = (1.29 atm) (85.9 L) P1V1P1V1 P2V2P2V2 The Combined Gas Law T1T1 T2T2 = X 0.299 atm=
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Ex. (2) If 82.7 L of xenon gas at 142.9 kPa & 50.3 o C were compressed to a volume of 28.3 L and then was returned to standard pressure, what would be its new Kelvin temperature? P1V1P1V1 P2V2P2V2 T1T1 T2T2 (101.3 kPa)(82.7 L) 323.3 KX = (142.9 kPa) (28.3 L) The Combined Gas Law = X 78.4 K=
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Ex. (3) If 3.75 L of chlorine gas at 20.0 o C and 105.0 kPa were returned to STP conditions, what would be the new volume? P1V1P1V1 P2V2P2V2 T1T1 T2T2 (101.3 kPa)(3.75 L) 293.0 K273.0 K (105.0 kPa) (X) The Combined Gas Law = X 3.62 L= =
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Ex. (4) If 25.0 L of a gas at STP were heated to a temperature of 39.0 o C and the volume expanded to 68.3 L, then what would be the new pressure in atm? P1V1P1V1 P2V2P2V2 T1T1 T2T2 (X)(25.0 L) 273.0 K312.0 K (1.00 atm) (68.3 L) The Combined Gas Law = P2P2 0.418 atm= =
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