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Question

14g oxygen at 0C and 10 atm are subjected to reversible adiabatic expansion to a pressure of 1 atm. What is the work done?

A
-12.80 liter atm
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B
-11.82 liter atm
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C
-11.90 liter atm
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D
-12 liter atm
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Solution

The correct option is B -11.82 liter atm

P1 = 10 atm at T = 273k for 1432 mole of O2

P2 = 1 atm at T = T2k for 1432 mole of O2

For adiabatic expansion we have Tγ P1 γ = constant

(T1T2)γ = (P2P1)1 γ

or γ = (T1T2) = (1 γ)log(P2P1)

or 1.4 log 273T2 = (1 1.4)log(110)

T2 = 141.4 k

Work done in adiabatic expansion = nR(γ 1)(T2 T1)

= 1432 × 0.0821(141.4 273)(1.4 1)

Wrev = 11.82 liter atm


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