It is interesting to note just how much has been already covered without explicitly mentioning the famous three laws of thermodynamics. It is very tempting to begin with mentioning the three laws and then studying thermal physics, after all supposedly everything banks on them. It is my opinion that this mode of proceeding while it works well for classical mechanics or quantum mechanics does not work well (pedagogically) for thermal physics. Rather than stating the laws and seeing the consequences as is done in classical mechanics with Newton's laws , it is far better in thermal physics to see how the laws emerge. Remember our goal is to find a more natural way of formulating the second law that fits well in Thermal physics and does not involve bringing in concepts that are better left in statistical mechanics.
We begin our discussion with an experimental fact that was discovered by Joule.
If a thermally isolated system is brought from one equilibrium state to another the work necessary to achieve this change is independent of the process used.
Note: Before we argued correctly that the amount of work done was path dependent. But for the special case of adiabatic work it is path independent.
So we are going from one equilibrium state to another and calculating some quantity \( W_{adiabatic}\) that is path independent (does not depend on its history). Recall, this was the special and lovable feature of state functions. Therefore this existence of this experimental fact implies the existence of a state function so that \( W_{adiabatic} = U_2 - U_1 = \Delta U \). The choice of using the letter U is supposed to be suggestive. Remember the system is thermally isolated and yet we are able to change it from one equilibrium state to another. This must mean that we are changing its internal energy.
But let us suppose that the system is not thermally isolated. Then as pointed out before, the work done on it depends on the path taken. Therefore we can imagine to sorts of scenarios: We have a system A and we done work on it to bring it from one equilibrium state ,a, to another,b, while it is thermally isolated to get \( \Delta U \) but we can also put it into thermal contact with its surrounding and perform enough work to change it from equilibrium state a, to equilibrium state b. We will not do the same amount of work for both these processes. The difference between these two is heat, Q i,e Q = \( \Delta U - W\). Rearranging, the variables we arrive at what is called the first law of thermodynamics
\begin{equation}
\Delta U = Q + W \hspace{10mm} \text{eq.1}
\end{equation}
Note that in contrast to other discussions we have not talked about whether the process is reversible or not ,whether it is quasi-static or not. This is what gives it its generality. Heat therefore is the non-mechanical exchange of energy between the system and the surroundings because of their temperature difference.
There is a subtle difference between what we mean by heat and what we mean by work, or rather what physically heat means as opposed to work. When we have heat put into a system we increase he random motion of the constituent molecules, but when we change the energy of a system by doing work we displace the molecules in an ordered way. From a quantum mechanical perspective, n particles can exist in discrete energy levels \( \epsilon_i \). Let us say that there are \( n_i \) particles on level \( \epsilon_i \). The total internal energy U = \( \sum n_i \epsilon_i \). When we do work we change the levels \( \epsilon_i \) with the populations remaining the same, but the populations \( n_i \) can be changed and this is heat.
We can now use the first law of thermodynamics to justify the statement that heat lost is heat gained. This is a crucial assumption or may a obvious assumption used in carnot cycles and engines.
Imagine we a system that we have an adiabatic system so that no heat comes in or goes out and this system and has two subsystems A and B. We can apply the first law to each subsystem separately as follows:
\begin{eqnarray}
U_f^A - U_i^A &=& Q^A + W^A \\
U_f^B - U_i^B &=& Q^B + W^B
\end{eqnarray}
We add the two equations to get
\begin{equation}
U_f^A + U_f^B - (U_i^A + U_i^B) = Q^A +Q^B + W^A + W^B
\end{equation}
The term of the left hand side of the above equation is the change in internal energy of the composite system. The first term on the right hand side is the net flow of heat in the composite system and the second term is the work done on the composite system. But remember our assumption was that the system was adiabatic so it must be that \(Q^A + Q^B = 0\) and therefore \( Q^A = -Q^B\).
We begin our discussion with an experimental fact that was discovered by Joule.
If a thermally isolated system is brought from one equilibrium state to another the work necessary to achieve this change is independent of the process used.
Note: Before we argued correctly that the amount of work done was path dependent. But for the special case of adiabatic work it is path independent.
So we are going from one equilibrium state to another and calculating some quantity \( W_{adiabatic}\) that is path independent (does not depend on its history). Recall, this was the special and lovable feature of state functions. Therefore this existence of this experimental fact implies the existence of a state function so that \( W_{adiabatic} = U_2 - U_1 = \Delta U \). The choice of using the letter U is supposed to be suggestive. Remember the system is thermally isolated and yet we are able to change it from one equilibrium state to another. This must mean that we are changing its internal energy.
But let us suppose that the system is not thermally isolated. Then as pointed out before, the work done on it depends on the path taken. Therefore we can imagine to sorts of scenarios: We have a system A and we done work on it to bring it from one equilibrium state ,a, to another,b, while it is thermally isolated to get \( \Delta U \) but we can also put it into thermal contact with its surrounding and perform enough work to change it from equilibrium state a, to equilibrium state b. We will not do the same amount of work for both these processes. The difference between these two is heat, Q i,e Q = \( \Delta U - W\). Rearranging, the variables we arrive at what is called the first law of thermodynamics
\begin{equation}
\Delta U = Q + W \hspace{10mm} \text{eq.1}
\end{equation}
Note that in contrast to other discussions we have not talked about whether the process is reversible or not ,whether it is quasi-static or not. This is what gives it its generality. Heat therefore is the non-mechanical exchange of energy between the system and the surroundings because of their temperature difference.
There is a subtle difference between what we mean by heat and what we mean by work, or rather what physically heat means as opposed to work. When we have heat put into a system we increase he random motion of the constituent molecules, but when we change the energy of a system by doing work we displace the molecules in an ordered way. From a quantum mechanical perspective, n particles can exist in discrete energy levels \( \epsilon_i \). Let us say that there are \( n_i \) particles on level \( \epsilon_i \). The total internal energy U = \( \sum n_i \epsilon_i \). When we do work we change the levels \( \epsilon_i \) with the populations remaining the same, but the populations \( n_i \) can be changed and this is heat.
We can now use the first law of thermodynamics to justify the statement that heat lost is heat gained. This is a crucial assumption or may a obvious assumption used in carnot cycles and engines.
Imagine we a system that we have an adiabatic system so that no heat comes in or goes out and this system and has two subsystems A and B. We can apply the first law to each subsystem separately as follows:
\begin{eqnarray}
U_f^A - U_i^A &=& Q^A + W^A \\
U_f^B - U_i^B &=& Q^B + W^B
\end{eqnarray}
We add the two equations to get
\begin{equation}
U_f^A + U_f^B - (U_i^A + U_i^B) = Q^A +Q^B + W^A + W^B
\end{equation}
The term of the left hand side of the above equation is the change in internal energy of the composite system. The first term on the right hand side is the net flow of heat in the composite system and the second term is the work done on the composite system. But remember our assumption was that the system was adiabatic so it must be that \(Q^A + Q^B = 0\) and therefore \( Q^A = -Q^B\).
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