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In this situation, I believe that the balloon filled with gas is expanding and thus the gas is doing work on the balloon (causing the expansion), but at the same time heat is given off by the balloon, and this is what is maintaining the same temperature. Remember when temperature is kept constant, we can say that delta(U)=0, and since delta(U)= w+q, it follows that w=-q. -q being the heat given off by the balloon, and +w being the work done on the balloon.
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