When 10 liters of a gas at 1 atm is compressed to 3 liters at constant temperature , what property of the gas changes ?

We know that at state 1, we can describe the gas at STP by P1V1=n1RT1
We change the volume, so now there is a second state, state 2 that can be described as
P2V2=n2RT2
We are told that temperatures are constant and since no gas is lost then moles don't change. Hence n1T1=n2T2. So rearrange the equations as follows
P1V1/R=n1T1
P2V2/R=n2T2
Set them equal per my explanation above:
P1V1/R=P2V2/R
Now multiply by R on both sides
P1V1=P2V2
We want to find the pressure at state 2
P1V1/V2=P2
(1 atm)(10L)/(3L) = 3.33 atm = P2

If an object was to absorb heat energy, then that means there would be an increase in temperature. An increase in temperature would not lead to freezing or condensation, so we can eliminate the choices B and D. 

By definition, deposition is the process by which sediment settles out of the water or wind that is carrying it and is deposited in a new location.

Since deposition has nothing to do with absorbing heat energy, we can eliminate choice C. 

The best answer is A. vaporization. Vaporization involves the phase change from liquid to a gas; there are types of vaporization such as boiling or evaporation.