A mono-atomic ideal gas of two moles is taken through a cyclic process starting from `A` as shwon in the figure below. The volume ratios are `V_(B)//V_(A) = 2` and `V_(D)//V_(A) = 4`. If the temperature `T_(A)` at `A` is `27^(@)C`. Calculate a. The temperature of gas at `B`. b. Heat absorbed or evolved in each process. c. Total work done in cyclic process.
A
`1200` cal
B
`3000` cal
C
`2500` cal
D
None of these
Text Solution
Verified by Experts
The correct Answer is:
A
Charles law : `V prop T , (V_(A))/(T_(A)) = (V_(B))/(T_(B)) = (T_(B))/(T_(A))` `T_(B) = 2 xx 300 = 600 K` `W_(cyclic) = W_(AB) + W_(BC) + W_(CD) + W_(DA)` `W_(AB)("isobaric") = -nRDeltaT` `= - 2 xx 2 xx (600 - 300)` `= - 1200 cal` `W_(BC) ("isothermal") = - 2.303 nRT log'(V_(2))/(V_(1))` `= -2.303 xx 2 xx 600 xx log'(2)/(1)` `= -1658.16 cal` `W_(CD) ("Isochoric") = 0` `W_(DA) ("isothermal") = - 2.303 xx 2 xx 2 xx300 xx log'(1)/(4)` `= +1658.16 cal` for cyclic process `DeltaE = 0` therefore from `FLOT Q = -W_(cyclic) = 1200 cal`
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