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Gas Laws

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Collisions between particles of a gas and the walls of the container cause the pressure in a closed container of gas Factors that affect the pressure of an enclosed gas are its temperature, its volume, and the number of its particles.

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Charles’s Law Charles's law states that the volume of a gas is directly proportional to its temperature in kelvins if the pressure and the number of particles of the gas are constant.

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Boyle’s Law Boyle's law states that the volume of a gas is inversely proportional to its pressure if the temperature and the number of particles are constant

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Combined Gas Law The combined gas law describes the relationship among the temperature, volume, and pressure of a gas when the number of particles is constant.

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**A gas has a volume of 5. 0 L at a pressure of 50 kPa**

A gas has a volume of 5.0 L at a pressure of 50 kPa. What happens to the volume when the pressure is increased to 125 kPa? The temperature does not change. P1=50 kPa V1= 5.0 L P2= 125 kpa V2=

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**(P1=50 kPa) * (V1= 5.0 L) = (P2= 125 kpa) *V2**

Solve for V2 (50*5)=5* V2 5

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**2. Gas stored in a tank at 273 K has a pressure of 388 kPa**

2. Gas stored in a tank at 273 K has a pressure of 388 kPa. The safe limit for the pressure is 825 kPa. At what temperature will the gas reach this pressure? T1=273 K P1=388 kPa T2= P2= 825 kPa

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P1=388 kPa P2= 825 kPa T1=273 K = T2=? cross multiply (P1=388 kPa) (T2)= (P2= 825 kPa) (T1=273 K)

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**3. At 10°C, the gas in a cylinder has a volume of 0. 250 L**

3. At 10°C, the gas in a cylinder has a volume of 0.250 L. The gas is allowed to expand to 0.285 L. What must the final temperature be for the pressure to remain constant? (Hint: Convert from degrees Celsius to kelvins using the expression °C + 273 = K.) V1=.250 L T1= = 283 K V2= .285 L T2

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V1=.250 L V2= .285 L T1283 K = T2 cross multiply (V1=.250 L)* (T2) = ( V2= .285 L)* (T1283 K)

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