U is internal energy, which is the sum of heat + work done.(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\Delta[/tex]U = q + w -----(1) where w = [tex]\Delta[/tex](PV)

Enthalpy is the internal energy + work done

[tex]\Delta[/tex]H = [tex]\Delta[/tex]U + P[tex]\Delta[/tex]V

At constant pressure,

[tex]\Delta[/tex]H = q - P[tex]\Delta[/tex]V + P[tex]\Delta[/tex]V

[tex]\Delta[/tex]H = q[tex]_{p}[/tex]

Where q[tex]_{p}[/tex] is used to denote heat energy under constant pressure.

I'm confused about a few things here.

1) What is the work that is done here?

- P[tex]\Delta[/tex]V > What work is done?

+ P[tex]\Delta[/tex]V > What work is done?

By canceling them out, does it mean that no work is done at all i.e no volume change?

2) If there is a change in the number of moles of gas, + P[tex]\Delta[/tex]V is actually has an impact on the equation. I'm under the impression that [tex]\Delta[/tex]U consists of thermal energy only(i.e no PV work done) when we're talking about enthalpy.

Similarly under constant volume, from equation 1,

[tex]\Delta[/tex]U = q[tex]_{v}[/tex] because [tex]\Delta[/tex]V = 0, hence w = 0.

Since [tex]\Delta[/tex]H = [tex]\Delta[/tex]U + [tex]\Delta[/tex](PV)

and [tex]\Delta[/tex]V = 0,

therefore [tex]\Delta[/tex]H = q[tex]_{v}[/tex]?

1) My next question is why is there this [tex]\Delta[/tex](PV) term here that doesn't do anything at all?

2) Like the previous case, I'm under the impression that[tex]\Delta[/tex]U = thermal energy when we're talking about [tex]\Delta[/tex]H = [tex]\Delta[/tex]U + [tex]\Delta[/tex](PV)?.

Of course which is not true since [tex]\Delta[/tex]U = q + w. So what exactly is the difference between the w in [tex]\Delta[/tex]U = q + w and [tex]\Delta[/tex](PV) in enthalpy equation [tex]\Delta[/tex]H = [tex]\Delta[/tex]U + [tex]\Delta[/tex](PV)?

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# Enthalpy vs Internal energy for first law

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