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A gas in a balloon expands from 2L to 3L against 2 atm of pressure, what is the change in temperature relative to the initial temperature?
Since the pressure is constant, V/T = constant
V1/T1 = V2/T2
T2/T1 = V2/V1 = 3/2
(T2-T1)/T1 = 3/2 - 1 = 1/2
It means that the temperature is increased by 1/2 of its initial temperature.
The general equation governing changes in fixed-quantity gasses is
--------- = -------
This is what is known as the "combined gas law" because it covers the other known laws with separate names:
- Boyle's Law - constant temperature - P1V1 = P2V2
- Gay-Lussac's Law - constant volume - P1T2 = P2T1
- Charles's Law - constant pressure - V1T2 = V2T1
In this case, since the pressure is constant (2 atm) but volume and temperature are changing, we can use Charles's Law:
V1T2 = V2T1
(2)(T2) = (3)(T1)
T2 = 3/2 T1
So the final temperature is 1.5x the initial temperature. Note specifically that both the initial and final temperature in this equation must be measured in Kelvins (absolute temperature) - you cannot use Fahrenheit or even Celsius, as they are not the natural temperature scales. If the problem gives initial temperatures in either °F or °C then you must convert to Kelvin first:
K = (5/9)(F - 32) - 273.15
= C - 273.15
where F is degrees Fahrenheit and C is degrees Celsius.