The general equation governing changes in fixedquantity gasses is
P_{1}V_{1} P_{2}V_{2}
 = 
T_{1} T_{2}
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  P_{1}V_{1} = P_{2}V_{2}
 GayLussac's Law  constant volume  P_{1}T_{2} = P_{2}T_{1}
 Charles's Law  constant pressure  V_{1}T_{2} = V_{2}T_{1}
In this case, since the pressure is constant (2 atm) but volume and temperature are changing, we can use Charles's Law:
V_{1}T_{2} = V_{2}T_{1}
(2)(T_{2}) = (3)(T_{1})
T_{2} = 3/2 T_{1}
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.
11/15/2012

Michael B.
Comments
You didn't answer "the change in temperature relative to the initial temperature".
I disagree.
At worst, I think that is a matter of interpretation on an ambiguous question, and a matter of semantics. Does "the change" refer to an additive change or a multiplicative change. Both are "relative to" the initial temperature. IMO, most questions of this sort expect a multiplicative factor, not an additive one.
I saw that you had posted an additive answer, so I chose to go with the other option. But even so, when you say "it increased by 1/2 of it's original temperature" that is no different than when I say "the final temperature is 1.5x the initial temperature". It's just you stated the change as an additive term, and I stated it as a multiplicative factor. It's the same thing, just the same as saying "my bank account went up by 50%" vs "my bank account is 1.5 time what it used to be".... the only difference is in the semantics, but it's the same answer.