Thermodynamics questions

1. Jun 10, 2012

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!

2. Jun 10, 2012

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

3. Jun 10, 2012

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. 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

4. Jun 10, 2012

tdod

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

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

5. Jun 10, 2012

Redbelly98

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

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

6. Jun 10, 2012

Staff: Mentor

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.

$\Delta$H= $\Delta$E + $\Delta(pV)$

7. Jun 10, 2012

Staff: Mentor

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.

8. Jun 10, 2012

ehild

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

Last edited: Jun 10, 2012