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Pressure and Temperature William Thomson “Lord Kelvin”

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Presentation on theme: "Pressure and Temperature William Thomson “Lord Kelvin”"— Presentation transcript:

1 Pressure and Temperature William Thomson “Lord Kelvin”

2 NM Standards

3 Measuring Pressure The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17 th century. The device was called a “barometer” Baro = weight Meter = measure

4 An Early Barometer The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.

5 Pressure  Is caused by the collisions of molecules with the walls of a container  is equal to force/unit area

6 Standard Pressure 1 standard atmosphere (atm)  1 standard atmosphere (atm)  101.3 kPa (kilopascals)  14.7 lbs/in2  760 mm Hg (millimeters of mercury)  760 torr

7 Pressure UnitSymbolDefinition/Relationship PascalPaSI pressure unit 1 Pa = 1 newton/meter 2 Millimeter of mercury mm HgPressure that supports a 1 mm column of mercury in a barometer AtmosphereatmAverage atmospheric pressure at sea level and 0  C Torrtorr1 torr = 1 mm Hg Pressure is the force created by the collisions of molecules with the walls of a container

8 Manometer

9 An open manometer is filled with mercury and connected to a container of hydrogen gas. The mercury level is 57 mm higher in the arm of the tube connected to the hydrogen. If the atmospheric pressure is 0.985 atm, what is the pressure of the hydrogen gas, in atmospheres? An open manometer connected to a tank of argon has a mercury level 83 mm higher in the atmospheric arm. If the atmospheric pressure is 76.9 kPa, what is the pressure of the argon in kPa?

10 Another problem An open manometer is filled with mercury. The mercury level is 12 mm higher on the side open to the atmosphere. What is the total pressure of the gas, in kPa, if the atmospheric pressure is 100.8 kPa?

11 The Kelvin Scale

12 Standard Temperature Standard Temperature equals:  273 Kelvin (273 K)  0  C

13 Converting Celsius to Kelvin Gas law problems involving temperature require that the temperature be in KELVINS! Kelvins =  C + 273 °C = Kelvins - 273

14 Standard Temperature and Pressure “STP” Either of these:  273 Kelvin (273 K)  0  C Either of these:  273 Kelvin (273 K)  0  C And any one of these: 1 atm  1 atm  101.3 kPa  14.7 lbs/in 2 (psi)  760 mm Hg  760 torr And any one of these: 1 atm  1 atm  101.3 kPa  14.7 lbs/in 2 (psi)  760 mm Hg  760 torr


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