U vs H
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Re: U vs H
Delta U is the change in internal energy, whereas delta H is the change in enthalpy. I believe delta H is the total energy change, whereas delta U only accounts for the kinetic and potential energy of a system.
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Re: U vs H
To add on, the equation we often use to relate the two is ∆U = q + w. At constant pressure, ∆H = qp. Thus, we can rewrite the equation as ∆U = ∆H - P∆V when pressure is constant.
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Re: U vs H
Moreover, we can find that deltaU = qv, which makes sense because the volume is constant (no expansion work is being done), so the change in internal energy must solely be a result of the increase of energy inputted into the system qv at a constant volume.
On the other hand, deltaH = qp, which can be derived from the equation relating both deltaU and deltaH. DeltaH = deltaU + PdeltaV, which means deltaH = q+w + PdeltaV. Since w = -PdeltaV, deltaH = q + -PdeltaV +PdeltaV, and since we're assuming that the pressure is constant from qp, we can cancel the last two terms to get deltaH = qp.
On the other hand, deltaH = qp, which can be derived from the equation relating both deltaU and deltaH. DeltaH = deltaU + PdeltaV, which means deltaH = q+w + PdeltaV. Since w = -PdeltaV, deltaH = q + -PdeltaV +PdeltaV, and since we're assuming that the pressure is constant from qp, we can cancel the last two terms to get deltaH = qp.
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Re: U vs H
The change in internal energy is a summation of the enthalpy and the work done on the system so Delta U is Delta H +w
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Re: U vs H
Delta U represents the change in internal energy while delta H represents the change in enthalpy.
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Re: U vs H
A note on delta U being expressed the change in internal energy; (U=w+q); w is often expressed interchangeably as work done on the system and work done BY the system. For the former, it is w+q; for the latter, it is actually -w+q because the work done by the system is still positive but it results in a decrease in U.
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Re: U vs H
Chris Tai 1B wrote:Moreover, we can find that deltaU = qv, which makes sense because the volume is constant (no expansion work is being done), so the change in internal energy must solely be a result of the increase of energy inputted into the system qv at a constant volume.
On the other hand, deltaH = qp, which can be derived from the equation relating both deltaU and deltaH. DeltaH = deltaU + PdeltaV, which means deltaH = q+w + PdeltaV. Since w = -PdeltaV, deltaH = q + -PdeltaV +PdeltaV, and since we're assuming that the pressure is constant from qp, we can cancel the last two terms to get deltaH = qp.
Is this something we are supposed to memorize, or can this be derived from the equations given to us?
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Re: U vs H
Delta U is change in internal energy, while delta H is change in enthalpy. Under certain conditions, they can be equal to each other.
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Re: U vs H
805317518 wrote:Chris Tai 1B wrote:Moreover, we can find that deltaU = qv, which makes sense because the volume is constant (no expansion work is being done), so the change in internal energy must solely be a result of the increase of energy inputted into the system qv at a constant volume.
On the other hand, deltaH = qp, which can be derived from the equation relating both deltaU and deltaH. DeltaH = deltaU + PdeltaV, which means deltaH = q+w + PdeltaV. Since w = -PdeltaV, deltaH = q + -PdeltaV +PdeltaV, and since we're assuming that the pressure is constant from qp, we can cancel the last two terms to get deltaH = qp.
Is this something we are supposed to memorize, or can this be derived from the equations given to us?
We don't necessarily have to memorize this, but it would be ideal if we understand how to derive it or explain why it is what it is. If we know how to derive the equation, then we'll naturally be able to memorize it over time.
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