# Adiabatic Expansion: Constant pressure and temperature

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?

Hi Hpatps1,
For the adiabatic expansion we have the relation between T and V as T1V11-$\gamma$=T2V21-$\gamma$.
Where T1,V1 and T2,V2 are data of the system at two different states. If T is constant then T1=T2. Then by the above equation V1=V2. Hence dV is zero, and the W. That's why ΔU is zero.
Now for the next. By the above data it is clear that dV is zero iff T is constant (ΔT=0). So in the equation ΔU=CvΔT, ΔU certainly becomes zero when ΔT=0.
Regards.

Thanks! That makes sense.

As a gas expands or contracts adiabatically, although q=0, the temperature changes as a result of the particles doing work on the surroundings, so there is a change in internal energy equal to the work done by or to the system. U is only dependent on the kinetic energies of the molecules of the gas in an ideal system.

Andrew Mason
Homework Helper
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?
I am not sure why you say W = 0. Why do you think W = 0?

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?
Why would ΔU = 0?

AM