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One mole of an ideal monoatomic gas is taken from state A to state B through the process `P=3/2 T^(1//2)` It is found that its temperature increases by 100 K in this process. Now it is taken from state B to C through a process for which internal energy is related to volume as `U= 1/2 V^(1//2)` Find the total work performed by the gas (in Joule) if it is given that volume at B is `100 m^(3)` and at C it is `1600m^(3)` [Use `R= 8.3 J //mol-K`] |
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Answer» Correct Answer - `435` Process A rightarrow B `W_(AB)=intPdv=int3/2T^(1//2)dv=int3/2T^(1//2)xx1/3RT^(-1//2)dT` On solving `W_(AB)=50R=50xx8.3=415`J Process B rightarrow C `U=1/2 V^(1//2)` `3/2RT=1/2V^(1//2) Rightarrow 3PV^(1//2)=1` `:.P=1/3sqrt(V)` Now `W_(BC) = intPdv=underset100overset1600int1/(3sqrt(V))dv=2/3 sqrt(V)=2/3[40-10]=2/3xx30=20` J Total `W=415+20=435]` |
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