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?

Question

This was all the information given.  Please help me solve this!  Thank you!

Answer 1

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.

Answer 2

The general equation governing changes in fixed-quantity gasses is

  P1V1          P2V2
---------  =  -------
   T1            T2

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.