1.

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.

Answer»

Work done in PROCESS AB is - 400 R
Change in intert, al energy in process CA is 800 R.
Heat transferred in the process BC is 2000 R
Change in internal energy in process CA is -900 R.

Solution :Process AB:PT=constant =k
`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


Discussion

No Comment Found

Related InterviewSolutions