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If the boundary of system moves by an in...

If the boundary of system moves by an infinitiesimal amount, the work involved is given by `dw= -P_("ext")` dV, for irreversible process `W= -P_("ext") Delta V` (where `DeltaV= V_(f)- V_(i)`). For reversible process `P_("ext") = P_("int") +- dP ~~ P_("int")`, so for reversible isothermal process `W= -nRT`ln `(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)= 8` 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 against a constant pressure of 20 bar until `P_("gas") = P_("ext")`.
Total work done on the gas in two stage compression is

A

40

B

80

C

160

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`w_("irr (total)")=w_(1)+w_(2)`
`=-10((nRT)/(10)-(nRT)/(2))`
`=-20((nRT)/(20)-(nRT)/(10))`
`=5xxnRT=80 " bar-L"`
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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") Order of magnitude work is

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:

Derive the equation for 'W_("rev")' in isothermal reversible process.

q=w=-p_(ext)(v_(f)-v_(i)) is for irreversible…. Change.