If the boundary of system moves by an infinitesimal amount, the work involved is given by $$dw=$$−Pext dVfor irreversible process $$w=$$−Pext ΔV (where$$ Δ$$V=Vf$$−Vi $$) for reversible process Pext $$=Pint$$ ±dP≃Pint so V for reversible isothermal process $$w=$$−nRTln⁡$$VfVi2$$ mole of an ideal gas undergoes isothermal compression along three different paths: different paths: (i) reversible compression from $$Pi=2$$ bar and $$Vi=8L$$ to $$Pf=20$$ bar (ii) a single stage compression against a constant external pressure of $$20$$ bar, and (iii) a two stage compression consisting initially of compression against a constant external pressure of $$10$$ bar until Pgas $$=Pext$$ , followed by compression against a constant pressure of $$20$$ bar until $$Pgas=Pext$$.Work done (in$$ $$bar-L) on the gas in reversible isothermal compression is: Work done on the gas in single stage compression isTotal work done on the gas in two stage compression is:

Question

If the boundary of system moves by an infinitesimal amount, the work involved is given by $$dw=$$−Pext dVfor irreversible process $$w=$$−Pext ΔV  (where$$ Δ$$V=Vf$$−Vi $$) for reversible process Pext $$=Pint$$ ±dP≃Pint so V for reversible isothermal process $$w=$$−nRTln⁡$$VfVi2$$ mole of an ideal gas undergoes isothermal compression along three different paths: different paths: (i) reversible compression from $$Pi=2$$ bar and $$Vi=8L$$ to $$Pf=20$$ bar  (ii) a single stage compression against a constant external pressure of $$20$$ bar, and (iii) a two stage compression consisting initially of compression against a constant external pressure of $$10$$ bar until Pgas $$=Pext$$ , followed by compression against a constant pressure of $$20$$ bar until $$Pgas=Pext$$.Work done (in$$ $$bar-L) on the gas in reversible isothermal compression is: Work done on the gas in single stage compression isTotal work done on the gas in two stage compression is:

Infinity Learn · Accepted Answer

$$PV=nRT$$⇒$$2$$×$$8=2$$×$$0.080$$×$$TT=100Kwrev$$ $$=$$−$$2.303$$×n×R×Tlog⁡$$P1P2=$$−$$2.303$$×$$2$$×$$0.08$$×$$100$$×log⁡$$110=36.848$$ $$bar-L$$ $$wirr=$$−$$PextV2$$−$$V1=$$−$$20nRTP2$$−$$nRTP1=144$$ $$bar-L$$ $$wirr($$ total $$)=w1+w2=$$−$$10nRT10$$−$$nRT2=$$−$$20nRT20$$−$$nRT10=5$$×$$nRT=80$$ $$bar-L$$

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