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 Δ V
(where Δ V = V f –V i )
For reversible process P ext = P int ± d p ≈ P int
So for reversible isothermal process
W = – nRT λ n 
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, and
(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 .
(i)Work done (in bar-L) on the gas in reversible
isothermal compression is –
Text Solution
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Ans.
(i)
Sol. PV = nRT ⇒ 2 × 8 = 2 × 0.080 × T
T = 100 K
W rev = –2.303 × n × R × T log 
= –2.303 × n × R × T × log 
= 2.303 × 2 × 0.08 × 100 × log 
= 36.848 bar - L
(ii)
Sol. W irr = – P ext (V 2 –V 1 )
= – 20 
= 144 bar–L
(iii)
Sol. W irr (total) = W 1 + W 2
= –10 
–20 
= 5 × nRT = 80 bar-L
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