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### Topic: Might be a dumb question, but why is deltaU=q+w?  (Read 541 times)

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#### humor5211

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##### Might be a dumb question, but why is deltaU=q+w?
« on: December 07, 2020, 07:13:23 AM »
In the beginning chapter of the Pchem textbook, it explains about the internal energy starts with
"All changes in a closed system in which no chemical reactions or phase changes occur can be classified only as heat, work, or a combination of both."

I wonder how did scientists end up with this conclusion. Did they just all agree to say this way bcz it makes sense with everything? How can we simply say deltaU=q+w ?

#### Beefsteak

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##### Re: Might be a dumb question, but why is deltaU=q+w?
« Reply #1 on: December 07, 2020, 07:25:15 PM »
So break this question down by first defining what a closed system is exactly. A closed system is defined as a physical system that does not allow transfer of matter in or out of the system. Alright so no matter can travel in or out of a closed system, so what does that leave us? In this universe by thermodynamics standards the only other option besides matter in the universe is energy. Also by thermodynamics standards, any energy in the most simplest terms can be expressed as some combination of heat (temperature difference between the system and surroundings) or some form of work weither it's mechanical, electrical or gravitational. Therefor the equation ΔU = q + w shows this relationship between the transfer of internal energy in a closed system and surroundings.

#### arslanali5050

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##### Re: Might be a dumb question, but why is deltaU=q+w?
« Reply #2 on: December 08, 2020, 01:31:58 AM »
There are only three ways we can increase its internal energy (which is the sum of its internal potential energy and internal kinetic energy): flow heat into it; do work on it; or add matter to it (and the converse for decreasing its internal energy). If the system is closed (no matter can flow in or out), we are left with only the first two, which naturally leads to the following equation:
ΔU=q+w
Now, suppose, It turns out that we can construct such a function by adding pV to U. We call that function the enthalpy, H:

H=U+pV=>ΔH=ΔU+Δ(pV)=ΔU+pΔV+VΔp

At constant pressure, Δp=0, so:

ΔHp=ΔU+pΔV=qp+wp+pΔV

But at constant p, psys=pext. Thus, if we only have pV work:
wp=−pextΔV=−pΔV

Hence:

ΔHp=qp−pΔV+pΔV=qp
Let's compare a reaction at constant pressure with one at constant volume. At constant V, with pV-work only, qV=ΔU (because, with pV-work only, wV=0). Suppose the reaction is exothermic, and the volume of the products is greater than that of the reactants, i.e., ΔV>0. You can see from the equations that ΔH will be less negative than ΔU, i.e., less heat is evolved at constant pressure.