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Hpatps1
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I'm new to the forum, so please be kind.
I was reading through my pchem textbook, and I noticed something. We're given the equation:
ΔU = q + w
For an adiabatic expansion, we're told that q = 0. Fair enough, no heat transfer. But when there is a constant T and change in V, my book says:
ΔU = 0 (U is constant)
I don't understand. W = -∫pdV, doesn't it? Why is work zero?
The book also says that for a perfect gas, U isn't dependent on volume. When they use this fact, it makes sense why ΔU = 0, but it seems like a contradiction when using the definition of work.
Also, the book continues and says:
ΔU = CvΔT when V is constant.
Again, W = -∫pdV, right? So dV = 0, shouldn't ΔU = 0?
Thanks in advance!
I was reading through my pchem textbook, and I noticed something. We're given the equation:
ΔU = q + w
For an adiabatic expansion, we're told that q = 0. Fair enough, no heat transfer. But when there is a constant T and change in V, my book says:
ΔU = 0 (U is constant)
I don't understand. W = -∫pdV, doesn't it? Why is work zero?
The book also says that for a perfect gas, U isn't dependent on volume. When they use this fact, it makes sense why ΔU = 0, but it seems like a contradiction when using the definition of work.
Also, the book continues and says:
ΔU = CvΔT when V is constant.
Again, W = -∫pdV, right? So dV = 0, shouldn't ΔU = 0?
Thanks in advance!