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If the boundary of system moves by an infinitesimal amount, the work involved is given by dw= -P_("ext") dV, for irreversible process W= -P_("ext") Delta V (where Delta V= V_(f) - V_(i)). For reversible process. P_("ext") = P_("int") +- dP ~= P_("int"), so for reversible isothermal process W = -nRT "In" (V_(f))/(V_(i)) 2 mole of an ideal gas undergoes isothermal compression along three different paths: (i) reversible compression from P_(i) = 2 bar and V_(i) = 8L " to" P_(f) = 20 bar (ii) a single stage compression against a constant external pressure of 20 bar (iii) a two stage compression consisting initially of compression against a constant external pressure of 10 bar until P_("gas") = P_("ext"), followed by compression aganist a constant pressure of 20 bar until P_("gas") = P_("ext") Work done (in bar -L) on the gas in reversible isothermal compression is: |
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Answer» <P>9.212 `= -2.303 xx 0.083 xx 2 xx ((P_(i)V_(i))/(NR)) ."LOG" (V_(F))/(V_(i))` `= -2.303 xx 0.083 xx 2 ((2 xx 8)/(2 xx 0.083)) "log" (2)/(20)` = 36.848 bar Lt |
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