[Physics] What’s the difference between work in thermodynamics and mechanics
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What is the difference between work in thermodynamics and work in mechanics?
Best Answer
Actually I think I disagree with the answer by BMS (the group of asymptotic symmetries of asymptotically flat spacetimes?). However I am not sure to have understood BMS'answer completely.
In my opinion, there is no difference between the definition of work in pure mechanics and work in thermodynamics (I stress that I am speaking of thermodynamics and not statistical mechanics). In both cases it is computed by the integral of ${\bf F} \cdot {\bf ds}$, taking all forces acting on the system into account. In the pure mechanical case the theorem of energy conservation says that
$$W = \Delta U + \Delta K\:.\qquad (1)$$
$W$ is the work done on the system by external systems, $K$ its kinetic energy and $U$ the total potential energy of internal forces. When considering situations where the work $W'$ of the system on the external systems coincides, up to the sign, to the work $W$ done by the external system on the system (and this is not the case discussed by BMS) we can also say that:
$$\Delta U + \Delta K + W' =0\:. \qquad (2)$$
In real physical systems, one has to consider the fact that a system receives energy also in terms of heat, $Q$: that is energy that cannot be described in terms of macroscopic work. In this case (1) has to be improved as
$$W + Q = \Delta U + \Delta K\:.\qquad (3)\:.$$
Actually, also the definition of $U$ has to be improved in (3), since it has to encompass the thermodynamic internal energy in addition to all types of macroscopic potential energies.
Referring to standard system of thermodynamics (thermal machines), where $\Delta K$ is negligible and the work done by the external system is identical up to the sign to that done by the system, (3) simplifies to
$$\Delta U = Q -W'\:,$$
that is the standard statement of the first principle of thermodynamics for elementary systems. However the general form is (3).
It is worth stressing that this picture needs a sharp distinction between macroscopic description (essentially done in terms of continuous body mechanics) and microscopic description, completely disregarded but embodied in the notions of heat and internal (thermodynamic) energy. If, instead one considers also the microscopic (molecular) structure of the physical systems, the distinction between work and heat is more difficult to understand since both are represented in terms of forces. Nevertheless exploiting the statistical approach to Hamiltonian mechanics the said distinction arises quite naturally.
Focusing on the system given by a rigid block discussed by BMS, the absolute value of the work $W$ done by the friction force acting to the block due to the ground (that eventually stops the block), is different from the absolute value of the work $W'$ done by the block on the ground. The former amounts to $W= -K$ the latter, instead, is $W' = 0$. The energy equation for the block is:
$$W + Q = \Delta U + \Delta K\:.$$
$Q$ is the non-mechanical energy entering the block during the process, responsible for the increase of its temperature. Since $W= -K$ one can simplify that equation to
$$Q= \Delta U\:.$$
The equation for the ground (for instance a table) is instead simply:
$$Q' = \Delta U'$$
Now $Q' \neq -Q$ and $W'=0 \neq -W$. The fact that $Q+Q' \neq 0$ it is important because it says that there is a heat source between the contact surfaces of the two bodies, and the total heat is not conserved (as conversely was supposed in the original theory of heat, the "flogisto" represented as a fluid verifying a conservation equation).
If referring to the overall system made of the block and the table, since no energy enters it, the equation is
$$\Delta U + \Delta U' + \Delta K =0\:.$$
That is
$$\Delta U + \Delta U' = -\Delta K >0\:.$$
It says that all the initial kinetic energy is finally transformed into internal energy producing the increase of temperature of both the block and the table.
Statistical Mechanics is the theory of the physical behaviour of macroscopic systems starting from a knowledge of the microscopic forces between the constituent particles.
The theory of the relations between various macroscopic observables such as temperature, volume, pressure, magnetization and polarization of a system is called thermodynamics.
In thermodynamics, work is the negative of the change in internal energy due to a change in volume, usually holding entropy and particle numbers constant. This takes the form of a force pushing on the walls of the volume, which connects it to our conventional notion of work, $W = F~\Delta x$, as seen for example if we consider a cylinder of cross-section $A$, $$W = F~\Delta x = F~\frac AA~\Delta x = \frac FA~A\Delta x = P ~\Delta V.$$And the change in internal energy is just the negative of the work, $-P~\Delta V,$ due to the law of energy conservation.
In fact we can also define that $W = P~\Delta V$ even when we are not holding entropy and particle numbers constant: but then it is not necessarily the same as the change in internal energy. So for example if you compress an ideal gas it generally heats up; you could still speak of the work as $P~\Delta V$ at constant temperature, but "at constant temperature" means essentially "we squeeze this thing and it wants to become warmer, but we let energy out of the system through the walls until it comes back to the same temperature": there has been a negative work, and perhaps the internal energy has still gone up, but it has not gone up as much as it would have had the walls been thermodynamic insulators. In these cases however we can often define a "free energy" (in this case $E - T S$) which the work is the negative of the change of.
Best Answer
Actually I think I disagree with the answer by BMS (the group of asymptotic symmetries of asymptotically flat spacetimes?). However I am not sure to have understood BMS'answer completely.
In my opinion, there is no difference between the definition of work in pure mechanics and work in thermodynamics (I stress that I am speaking of thermodynamics and not statistical mechanics). In both cases it is computed by the integral of ${\bf F} \cdot {\bf ds}$, taking all forces acting on the system into account. In the pure mechanical case the theorem of energy conservation says that $$W = \Delta U + \Delta K\:.\qquad (1)$$ $W$ is the work done on the system by external systems, $K$ its kinetic energy and $U$ the total potential energy of internal forces. When considering situations where the work $W'$ of the system on the external systems coincides, up to the sign, to the work $W$ done by the external system on the system (and this is not the case discussed by BMS) we can also say that: $$\Delta U + \Delta K + W' =0\:. \qquad (2)$$ In real physical systems, one has to consider the fact that a system receives energy also in terms of heat, $Q$: that is energy that cannot be described in terms of macroscopic work. In this case (1) has to be improved as $$W + Q = \Delta U + \Delta K\:.\qquad (3)\:.$$ Actually, also the definition of $U$ has to be improved in (3), since it has to encompass the thermodynamic internal energy in addition to all types of macroscopic potential energies.
Referring to standard system of thermodynamics (thermal machines), where $\Delta K$ is negligible and the work done by the external system is identical up to the sign to that done by the system, (3) simplifies to $$\Delta U = Q -W'\:,$$ that is the standard statement of the first principle of thermodynamics for elementary systems. However the general form is (3).
It is worth stressing that this picture needs a sharp distinction between macroscopic description (essentially done in terms of continuous body mechanics) and microscopic description, completely disregarded but embodied in the notions of heat and internal (thermodynamic) energy. If, instead one considers also the microscopic (molecular) structure of the physical systems, the distinction between work and heat is more difficult to understand since both are represented in terms of forces. Nevertheless exploiting the statistical approach to Hamiltonian mechanics the said distinction arises quite naturally.
Focusing on the system given by a rigid block discussed by BMS, the absolute value of the work $W$ done by the friction force acting to the block due to the ground (that eventually stops the block), is different from the absolute value of the work $W'$ done by the block on the ground. The former amounts to $W= -K$ the latter, instead, is $W' = 0$. The energy equation for the block is:
$$W + Q = \Delta U + \Delta K\:.$$
$Q$ is the non-mechanical energy entering the block during the process, responsible for the increase of its temperature. Since $W= -K$ one can simplify that equation to
$$Q= \Delta U\:.$$
The equation for the ground (for instance a table) is instead simply:
$$Q' = \Delta U'$$
Now $Q' \neq -Q$ and $W'=0 \neq -W$. The fact that $Q+Q' \neq 0$ it is important because it says that there is a heat source between the contact surfaces of the two bodies, and the total heat is not conserved (as conversely was supposed in the original theory of heat, the "flogisto" represented as a fluid verifying a conservation equation).
If referring to the overall system made of the block and the table, since no energy enters it, the equation is
$$\Delta U + \Delta U' + \Delta K =0\:.$$
That is
$$\Delta U + \Delta U' = -\Delta K >0\:.$$
It says that all the initial kinetic energy is finally transformed into internal energy producing the increase of temperature of both the block and the table.