prove:magnitude of work involved inan isothermal expansion is greater than that involved inan adiabatic exp.?

Question

thermodynamics(first law),adiabatic process(q=0),isothermal process(temp. difference=0)

Answer 1

A picture would be worth a thousand words, but I don't know how to draw this. If you plot P against V, then the work done is the area under the curve. If you'll excuse a new term, the term I refer to as "g" is really the greek letter gamma, which is the ratio of Cp to Cv, which is always greater than 1. In isothermal expansion PV = a constant, whereas in adiabatic expansion PV^g = a constant. So if both processes start off at the same P and V, then in the plot of P against V, the adabiatic plot "falls off" faster because g is greater than 1. Thus the area under the curve is smaller for the adabiatic expansion.

Answer 2

Well I gotta say that in last year's All-Star Game Soriano hit the ball into the booth looking part of the Giants stadium (I think its in Right Center field) and that Home Run was totally over looked by everyone and all evreyone was thinking about was Ichiro's ITP Homer and LaRussa's terrible management by not putting in Pujols the whole game even when he needed him I think that that Home Run would have been remembered by everyone if would have noticed exactly where that ball landed. That must be super hard to do especially for a right handed hitter

Answer 3

isothermal process work done by addition of extra heat from eviroment taking temperture constant;(internal energy=constant)

BUT adiabatic process(q=0) ie,total heat given during process=0 hence system spent some of internalenergy toexpand greter volume;

Answer 4

i cant ,give me zero mrks