3.2 Work, heat, and internal energy  (Page 2/6)

The expansion is isothermal, so T remains constant over the entire process. Since n and R are also constant, the only variable in the integrand is V , so the work done by an ideal gas in an isothermal process is

Notice that if $V_2>V_1$ (expansion), W is positive, as expected.

The straight lines from A to B and then from B to C represent a different process. Here, a gas at a pressure $p_1$ first expands isobarically (constant pressure) and quasi-statically from $V_1\text{to}{V}_2$ , after which it cools quasi-statically at the constant volume $V_2$ until its pressure drops to $p_2$ . From A to B , the pressure is constant at p , so the work over this part of the path is

From B to C , there is no change in volume and therefore no work is done. The net work over the path ABC is then

A comparison of the expressions for the work done by the gas in the two processes of [link] shows that they are quite different. This illustrates a very important property of thermodynamic work: It is path dependent . We cannot determine the work done by a system as it goes from one equilibrium state to another unless we know its thermodynamic path. Different values of the work are associated with different paths.

Internal energy

The internal energy     $E_{\text{int}}$ of a thermodynamic system is, by definition, the sum of the mechanical energies of all the molecules or entities in the system. If the kinetic and potential energies of molecule i are $K_i$ and $U_i,$ respectively, then the internal energy of the system is the average of the total mechanical energy of all the entities: