Is ∆E = q always true at constant volume?

[tdod]

True or False:
"The Heat, q, at constant volume is equal to the change in internal energy, ∆E, for the process at constant pressure."

I know that this is true, but i can't for the life of me figure out why. Can anyone explain?

Thanks!

 

[ehild]

Start from the First Law of Thermodynamics. You can change the internal energy by adding heat or/and doing work on the gas. How is work related to change of volume at constant pressure?

ehild

 

[Simon Bridge]

At constant volume, all the heat goes into internal energy - because it has nowhere else to go - but the pressure has to go up ... because of all that extra motion. So I'm a bit confused about what it is you are asserting is true. [edit]saw ehild's responce ... I think I get it.

The way to resolve confusions in thermodynamics is to concentrate on what each of the things are ... what is heat, and internal energy, and how does the gas give rise to pressure and temperature?
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cvpro.html
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cppro.html

 

[tdod]

Start from the First Law of Thermodynamics. You can change the internal energy by adding heat or/and doing work on the gas. How is work related to change of volume at constant pressure?

ehild

So, I know that at constant volume ∆E = q.

However, at constant pressure ∆E = q + w. Therefore, shouldn't they be different?

 

[Redbelly98]

... However, at constant pressure ∆E = q + w ...

Actually, that relation is true for any process, not just at constant pressure.

What do you know about w in a constant volume process?

 

[Chestermiller]

At constant volume, the change in internal energy is equal to the amount of heat added.

At constant pressure, the change in enthalpy is equal to the amount of heat added.

\DeltaH= \DeltaE + \Delta(pV)

 

[Chestermiller]

At constant volume, the change in internal energy is equal to the amount of heat added.

At constant pressure, the change in enthalpy is equal to the amount of heat added.

 

[ehild]

So, I know that at constant volume ∆E = q.

However, at constant pressure ∆E = q + w. Therefore, shouldn't they be different?

In general, ∆E = q + w, the internal energy changes with heat transfer and with work. But the work done by the gas is w=-pΔV for a very small change when the pressure can be considered constant. In general, w=-∫pdV where the integration goes between the initial and final volume. At constant volume, the initial and final volumes are the same, ΔV=0, so w=0. The change of internal energy at constant volume is equal to the heat added: ∆E = q.

ehild