Created by Alejandra Gomez
over 4 years ago
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PRESSURE EQUIVALENCES FOR GAS LAWS 1atm = 760mmHg= 760torr = 1.013 x 105 Pa = 101.3 KPa 1atm = 14.69psi or 1psi = 0.068atm 1 atm = 101325 Pa Pa refers to Pascals 1 atm = 1,01325 bar They are about the same, sometimes they refer to 1 atm as 1 bar mmHg refers to milimeters of Mercury ( column ) or Torricelis, another pressure measurement. p.s.i refers to " Pound per squeare inch " as the manometers usually use this pressure unit.
WHY IS THE KELVIN TEMPERATURE (°K) SCALE USED IN GAS LAWS? Jacques Charles ( 1746 - 1823 ) studied the effect of temperature on the volume of a gas, and he found out that the volume of any gas at 0°C, and constant pressure will increase by 1/ 273 for each 1°C increase in temperature. It would increase by 25/ 273 if the temperature increased 25°C. Not only the volume increased with the temperature increase, it increased by a predictable amount. Lowering the temperature would reduce the volume of a gas. If the temperature of 100L of gas at 0°C were lowered by 50°C, the volume of the gas would contract by 50/ 273 of the original volume. The new volume would be 81.7L ( 100L - 18.3L). What would happen to the volume of the gas if the temperature were lowered to - 273°C? Theoretically the volume should lower to zero. But this does not really happen as gases liquiefy first before reaching a temperature of -273°C. Yet the idea led William Thomson a British physicist with the title of Lord Kelvin, in 1848 to propose an absolut scale of temperature, which lowest temperature value was zero (absolute zero) that corresponded to a temperature of -273°C. There were no temperatures below absolute zero, so all temperatures above the absolute zero were positive. This absolute scale came to be known as the Kelvin scale (°K) . The principal difference between the Celcius and the Kelvin scales is the location of the zero temperature. The Kelvin temperature scale is used in all equations concerning volumes of gases and is part of the Charle's law. To convert from °C to Kelvin (°K ) all you have to do is add 273° ( °K = °C + 273°K) Example : 5°C are equivalent to 278°K ( 5°C + 273 = 278°K ) Taken from Chemistry Cliffs study solver pg 300
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