Heat Transfer at Constant Volume/Pressure

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Lisa Tang 1C
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Heat Transfer at Constant Volume/Pressure

Postby Lisa Tang 1C » Wed Jan 17, 2018 9:32 am

Can someone explain to me or clarify the following statement made in the textbook?
"At constant volume, the heat transfer is interpreted as delta U; at constant pressure, it is interpreted as delta H."
Why is this true? I think I partly understand that when volume doesn't change, no expansion work is being done, but I'm not sure.

Deap Bhandal L1 S1J
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Re: Heat Transfer at Constant Volume/Pressure

Postby Deap Bhandal L1 S1J » Wed Jan 17, 2018 10:07 am

I believe you are correct. At a constant pressure, Delta H = Delta U - PV and when both pressure and volume are constant, Delta H = Delta U.

Michelle Nguyen 2L
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Re: Heat Transfer at Constant Volume/Pressure

Postby Michelle Nguyen 2L » Wed Jan 17, 2018 5:00 pm

As the textbook assumes no non-expansion work is done, and deltaU= heat(q) + work(w), when the volume is constant throughout the reaction, this means no expansion work is done and so w= 0 and deltaU= q. The definition of deltaH is heat transferred at constant pressure, and takes into account that energy is also exchanged due to expansion work (change in volume), also assuming no non-expansion work is done. At constant pressure, w would likely have a nonzero value, and so deltaU no longer equals just q; instead, we use deltaH for energy transfer in the form of heat only. Therefore, heat transfer is deltaH at constant pressure, and deltaU at constant volume.

William Satyadi 2A
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Re: Heat Transfer at Constant Volume/Pressure

Postby William Satyadi 2A » Wed Jan 17, 2018 5:04 pm

The way I learned this in the past during physics was by looking at the graphs. If you look at an isobaric system (pressure is constant), the work done by the system can be calculated by finding the area under the graph. If you look at an isochoric system (constant volume), you'll see that the line moves only up and down, and there is no area under the graph. This is because work is force applied over a certain distance, and at a constant volume, there is no distance moved. Therefore, the work would be zero, and according to the first law, delta U=q. The link down below has some pretty good diagrams!

https://physics.info/pressure-volume/

Janine Chan 2K
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Re: Heat Transfer at Constant Volume/Pressure

Postby Janine Chan 2K » Wed Jan 17, 2018 8:49 pm

Here is a mathematical approach in addition to the responses above...
At constant volume, the heat transfer is interpreted as ∆U. This is true because ∆U = q + w, and w = -Pex∆V, where P is external pressure and V is volume. If the volume is constant, ∆V=0, which implies w=0, and thus ∆U=q.
At constant pressure, it is interpreted as ∆H. This is true because ∆H = ∆U + P∆V at constant pressure. Using ∆U = q + w and w = -Pex∆V, we can plug in ∆H = (q -Pex∆V) + P∆V. Note: In the textbook, it says that "because the system is open to the atmosphere or in a vessel that can adjust its size (because chemical reactions usually take place at constant pressure in vessels open to the atmosphere), the pressure of the system is the same as the external pressure; so Pex=P". Simplified, this equals ∆H=q.


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