One mole of a diatomic ideal gas `(gamma=1.4)` is taken through a cyclic process starting from point A. The process `AtoB` is an adiabatic compression, `BtoC` is isobaric expansion, `CtoD` is an adiabatic expansion, and `DtoA` is isochoric. The volume ratios are `V_A//V_B=16 and V_C//V_B=2` and the temperature at A is `T_A=300K`. Calculate the temperature of the gas at the points B and D and find the efficiency of the cycle.

Answer

A, B, D

Explanation

`n=1, for diatomic gas,
`gamma=1+2/5=7/5=1.4`
`AtoB,` adiabatic compression
`BtoC,` isobaric expansion
`CtoD,` adiabatic expansion
`DtoA,` isochoric process

Given `(V_A)/(V_B)=16,(V_C)/(V_B)=2`
`T_A=300K,T_B=?,T_D=?,eta=?`
For adiabatic compression process `AtoB`
`T_AV_A^(gamma-1)=T_BV_B^(gamma-1)` or
`T_B=((V_A)/(V_B))^(gamma-1) T_A=(16)^(2//5)xx300=909K`
`:.` For isobaric process`BtoC`: According to Charl's law
As `(V_B)/(T_B)=(V_C)/(T_C) or T_C=T_B((V_C)/(V_B))=902[2]=1818K`
For adiabatic expansion process `CtoD:`
As `(V_A)/(V_B)=16 and (V_C)/(V_B)=2, hence (V_A)/(V_C)=8`
According to Poisson's law,
`T_CV_C^(gamma-1)=T_DV_D^(gamma-1)`
`:. T_D=T_C[(V_C)/(V_D)]^(gamma-1)=1818[1/8]^(2//5)=1818/(64)^(1//5)=791K`
For `BtoC` process : Heat absorbed
`Q_1=nC_p(T_C-T_B)`
`=n(gammaR)/(gamma-1)(T_C-T_B)=1((7//5)R)/(2//5)(1818-909)`
`=(7R)/2xx909~=3182R`
For `DtoA`Process: Heat released
`Q_2=nC_v(T_D-T_A)=nR(gamma-1)(T_D-T_A)`
`=1.R/(2//5)(791-300)=-(5R)/2xx491`
`(:' No heat is exchanged in adiabatic processes).`
Now, `W_(AB)=-(nR)/(gamma-1)(T_B-T_A)`
`=-R/(2//5)(900-300)=-(5R)/2xx609`
`W_(BC)=-nR(T_C-T_B)=1xxR(1818-909)=909R`
`W_(CD)=-(nR)/(gamma-1)(T_C-T_D)=+R/(2//5)(1818-791)`
`=(5R)/2xx1027`
`W_(net)=909R+(5R)/2(1027-609)=909R+(5R)/2xx418`
`=909R+1045R=1954R`
`:. Efficiency =100xx(W_(net)//(Q_1))=100xx(1954R)/(3184R)=61.4%`