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From the first law, we have that $$dq = c_v dT+pd\alpha\tag{1},$$ that is, the total energy (by unit mass, where $\alpha$ is the specific volume) equals the variation in internal energy plus the work received/done by the system, in terms of the specific heat at constant volume. However, I struggle to understand the alternative form. I know that, mathematically we can write $$dq = c_v dT+d(p\alpha)-\alpha dp = c_pdT - \alpha dp \tag{2}.$$ What I don't understand is, how do we interpret this negative sign? What is the implication of the work term having the opposite sign in this form of the equation?

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  • $\begingroup$ Your first equation makes no sense to me. First law closed system $du=\delta q-\delta w$ or alternative (chemistry version) $du=\delta q +\delta w$. $\endgroup$
    – Bob D
    Mar 31, 2020 at 21:09
  • $\begingroup$ in this case, the definition of c$_v$ = dq/dT = du/dT (because volume is constant and therefore dw=0). Work is expressed as pressure times volume variation, leading to dq = c$_v$dT + pd$\alpha$. $\endgroup$
    – paulo
    Mar 31, 2020 at 21:18

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Your equations will confuse everyone. First, you're using $q$ for "total energy" whereas the letter $q$ normally refers to heat, and "total energy" is usually internal energy, $u$. Then you talk about $c_{v}dT$ as the "variation in internal energy". But $c_{v}dT$ is the change in internal energy only for an ideal gas. All of these will only confuse those who wish to help.

All of that aside, the business of the negative sign can be explained, using the mainstream equations for internal energy, by the fact that alternative forms of the first law are used in engineering and physics, versus chemistry. The form for engineering and physics is

$$\Delta U=Q-W$$

In chemistry it is often

$$\Delta U=Q+W$$

For the first version, work done by the system on the surroundings, meaning the system spends energy doing work, is given a positive value. That makes the internal energy of the system decrease, which makes sense since energy is extracted from system to perform work on the surroundings.

In the second, chemistry version, work done by the system on the surroundings is given a negative value, instead of a positive value. But it still makes sense since it also results in a decrease in internal energy, like the first version.

Both versions are consistent as long as you use the applicable sign for the value of work.

Hope this helps.

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