Gas Law Problems- Gay-Lussac's Law


Abbreviations				Conversions
atm - atmosphere			K  =  °C  +  273
mm Hg - millimeters of mercury		1 cm3 (cubic centimeter)  =  1 mL (milliliter)
torr - another name for mm Hg		1 dm3 (cubic decimeter)  =  1 L (liter)  =  1000 mL
Pa - Pascal (kPa  =  kilo Pascal)	Standard Conditions
K - Kelvin				0.00 °C  =  273 K
°C - degrees Celsius			1.00 atm  =  760.0 mm Hg  =  101.325 kPa  =  101,325 Pa

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56. Determine the pressure change when a constant volume of gas at 1.00 atm is heated from 20.0 °C to 30.0 °C.

57. A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at standard temperature?

58. A gas has a pressure of 699.0 mm Hg at 40.0 °C. What is the temperature at standard pressure?

59. If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what final pressure would result if the original pressure was 750.0 mm Hg?

60. If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0 atmospheres and its original temperature was 25.0 °C, what would the final temperature of the gas be?

61. A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased?

62. A sample of gas at 3.00 x 103 mm Hg inside a steel tank is cooled from 500.0 °C to 0.00 °C. What is the final pressure of the gas in the steel tank?

63. The temperature of a sample of gas in a steel container at 30.0 kPa is increased from -100.0 °C to 1.00 x 103 °C. What is the final pressure inside the tank?

64. Calculate the final pressure inside a scuba tank after it cools from 1.00 x 103 °C to 25.0 °C. The initial pressure in the tank is 130.0 atm.

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