What is the difference between the work done by a gas and the workflow that the same gas performs to exist? why does the variation in the internal energy of the equation dH = dQ + nRT not consider the work done by the gas to exist (working flow) but consumes only the heat transfer?
In the question below, the author considers dU = Q. He performs the calculation of the work done by the gas, separated from the internal energy (with an equation dH = dU + nRT ). If you know that in systems with constant volume dU = Q, but the gas volume varies (From 6 moles to 12 moles of gas), this does not imply an internal energy variation equal to dU = w + q (As stated in the 1st law of thermodynamics) and why consider dU = w + q and not dU = Q **? In practice or in what he did, the logic he did not use was.
Question:
“A constant volume calorimeter showed that the fuel loss in burning 1 mole of glucose is equal to -2559kJ in 298K, that is dU = -2555KJ”