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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 ${V}_{1}\phantom{\rule{0.2em}{0ex}}\text{to}\phantom{\rule{0.2em}{0ex}}{V}_{2}$ , 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.
where a and b are two parameters for a specific gas. Suppose the gas expands isothermally and quasi-statically from volume ${V}_{1}$ to volume ${V}_{2}.$ How much work is done by the gas during the expansion?
Because T is constant under the isothermal condition, the work done by 1 mol of a van der Waals gas in expanding from a volume ${V}_{1}$ to a volume ${V}_{2}$ is thus
Check Your Understanding How much work is done by the gas, as given in [link] , when it expands quasi-statically along the path ADC ?
${p}_{2}({V}_{2}-{V}_{1})$
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:
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