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One mole of monoatomic gas istaken through cyclic processas shown in the diagram. T_A=300 K. PL Process AB is defined PT=constant. Select the correct statements. |
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Answer» Work done in PROCESS AB is - 400 R `rArr (nRT^2)/V=k rArr (dV)/(dT)=(2nRT)/k` `W_(AB)=int_A^B PdV=int_A^B k/T dV` `W_(AB)=int_(T_A)^(T_B)k/T. (2nRT)/k dT` Now, `P_A/P_B=T_B/T_A rArr 1/3 =T_B/T_A rArr T_B=300/3`=100 K `therefore W_(AB)=int_300^100 2nR dT=2nR(100-300)` `rArr W_(AB)=-400nR=-400R` (`because` n=1 mole) Process CA:P/T =constant `therefore T_A//T_C = P_A//P_C , T_A//T_C=P_A//P_B` `rArr T_A//T_C=1//3 rArr T_C=3T_A` `T_C` = 900 K , `DeltaU=nC_V DeltaT` `=(1)3/2Rxx(T_A-T_C)=3/2Rxx(300-900)`=-900 R Process BC : Isobaric , `Q=nC_P DeltaT` `therefore Q=(1)5/2 Rxx(T_C-T_B)` `Q=5/2R xx (900-100)=5/2Rxx800`=2000 R |
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