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We got a question in a test, in which we were asked which system has zero change in internal energy and it had an option which was combustion of methane at constant temperature. I imagined this to be a situation in which the combustion is carried out in a system, where the change in internal energy of system due to heat released, is cancelled by the work done by the system. The reason why I thought of this is because of the formula, ∆U=nCv∆T, But this was wrong. When I asked my sir how this was wrong, he was not able to say why the ans using above stated formula is wrong. He tried giving me a logical explanation in which he said that a small part of the system releases heat which is used by the remaining part for it's combustion, thus the total temperature remains constant. but I am unable to understand why we choose to only study a part of the system while we have the entire system. It makes no sense to me. Could someone please explain how my sir's explanation is valid here and why the formula is wrong to use here?

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The change in internal energy of a reaction mixture is a function not only of the temperature change but also of the changes in the amounts of reactants and products in the mixture. This is because energy is consumed and released when we break- and make chemical bonds. The equation you wrote for the internal energy change applies only to a pure species or to a mixture of constant chemical composition. The occurrence of chemical reactions causes a change in the internal energy even at constant temperature.

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  • $\begingroup$ So you mean, the equation I wrote isn't suitable to be used for chemical reactions? One more thing, so can I say that when a chemical reaction occurs, even at constant temperature, it always has some enthalpy of reaction associated to it, so whenever there is change in enthalpy, there is change in internal energy using the relation, ∆H=∆U+P∆V? $\endgroup$ – Krish vasa Jan 2 at 16:13
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    $\begingroup$ Yes and yes to your questions. $\endgroup$ – Chet Miller Jan 2 at 16:40
  • $\begingroup$ Thanks a lot sir. $\endgroup$ – Krish vasa Jan 3 at 4:59

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