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Please explain and show work, thank you.

Image for 9. Shown here is a PV graph of a gas expanding (negative work), I can see that the area under the curve, the w

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Here we will compare a reversible vs. an irreversible expansion of the gas. Initially, lets think that the internal and external pressures are the same (Pint Pext). To expand the gas reversibly, we cut down the external pressure very slowly (technically, infinitely slowly) so the system never strays significantly from equilibrium. This is a reversible process because a very small increase in the external pressure would cause the gas to be compressed. To expand the gas irreversibly, we reduce the pressure very quickly. As the gas expands it does work on the surroundings (by moving the piston against the external pressure) The work done in these two processes is the area under the curves. (Remember that PV has units of energy or work). Clearly, more work is done by the reversible expansion. The reversible expansion does the maximum amount of work because the gas is pushing against the maximum possible external pressure. (Since the process remains at equilibrium throughout the reversible expansion, Pext -Pint) If the external pressure was any higher, the process would reverse and the gas would be compressed In general, energy released by a reversible process can do the maximum amount of work because less of the energy is lost as heat. A process that is done quickly (irreversibly) tends to generate turbulence and friction resulting in heat loss to the surroundings. Since in irreversible expansion pressure against surrounding is zero Since we know that w--Ap. Av Since in vacuum for expansion ??) will zero So w--Ap x Av (since Ap will be 0) 0 x Av W- eaternal force x distance through which piston moves Force (Area of cross section of piston x distance through which piston moves) Area of cross section of piston IfPm remains constant during the volume change then w=-Pet (V-v)

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