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Since the system is isothermal, we know that the change in internal energy is 0. So q+w=0, and q=-w. Since the balloon is expanding its volume against a changing pressure, work is being done. This also means that the energy lost during work is gained as heat, because the internal energy must stay 0. So heat is transfered in the system.
The question asked if work was done. We know this is true for this situation because W=-PdeltaV. Since the volume of the balloon decreased, deltaV is negative, which shows that work was done on the system (the balloon).
The isothermal reaction tells us that internal energy is equal to 0, which means q = -w. Since the temperature stayed constant while there was work being done, we know that there was a heat transfer from the surroundings into the system to maintain the temperature.
Temperature is increasing because the volume and pressure is changing in the balloon. The temperature may be the same but because of the change in P and V a little bit goes in and out so the temperature in the end is no change.
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