} p(V) \mathrm{d}V = -nRT_2 \ln \frac{V_2}{V_1}\\
\Delta W_{23} = \Delta W_{41} = 0\\
\Delta W_{34} = -nRT_1 \ln \frac{V_1}{V_2}
$$
On the isothermal curves the change in inner energy $\Delta U = \Delta W + \Delta Q$ is zero.
$$
\Delta Q_{12} = - \Delta W_{12} > 0\\
\Delta Q_{34} = - \Delta W_{34} < 0
$$
On the isochoric (isovolumetric) curves the heat quantities are
$$
\Delta Q_{23} = C_V (T_1 - T_2) < 0\\
\Delta Q_{41} = C_V (T_2 - T_1) > 0 
$$
The efficiency is then
$$
\eta = \frac{-\Delta W}{\Delta Q}
$$
$ \Delta Q$ is the input heat, i.e. sum of all the heat quantities $> 0$:
$$
\Delta Q = Q_{12}+Q_{41} =  n R T_2 \ln \frac{V_2}{V_1} + C_V (T_2 + T_1)
$$
$\Delta W$ is the total mechanical work:
$$
\Delta W = W_{12}+\Delta W_{34} = - nR(T_2 - T_1) \ln \frac{V_2}{V_1}
$$
So finally the efficiency is
$$
\eta = \frac{T_2 - T_1}{T_2 + \frac{C_V (T_2 - T_1)}{nR \ln V_2 / V_1}} < \eta_\text{C}.
$$
It is smaller than the efficiency of the Carnot cycle. But it should be equal to it if all processes are done reversibly.
The calculations are taken from a textbook (Nolting: Grundkurs Theoretische Physik 4) which actually points out this problem as a question to the reader. My only explanation is that this process is not reversible but I don't know how to tell without actually seeing how the
} p(V) \mathrm{d}V = -nRT_2 \ln \frac{V_2}{V_1}\\ 
 \Delta W_{23 } = \Delta W_{41 } = 0\\ 
 \Delta W_{34 } = -nRT_1 \ln \frac{V_1}{V_2 } 
 $ $ 
 On the isothermal curves the change in inner energy $ \Delta U = \Delta W + \Delta Q$ is zero. 
 $ $ 
 \Delta Q_{12 } = - \Delta W_{12 } > 0\\ 
 \Delta Q_{34 } = - \Delta W_{34 } < 0 
 $ $ 
 On the isochoric (isovolumetric) arch the heating system quantities are 
 $ $ 
 \Delta Q_{23 } = C_V (T_1 - T_2) < 0\\ 
 \Delta Q_{41 } = C_V (T_2 - T_1) > 0 
 $ $ 
 The efficiency is then 
 $ $ 
 \eta = \frac{-\Delta W}{\Delta Q } 
 $ $ 
 $ \Delta Q$ is the input heat, i.e. union of all the heat quantities $ > 0 $: 
 $ $ 
 \Delta Q = Q_{12}+Q_{41 } =   normality gas constant T_2 \ln \frac{V_2}{V_1 } + C_V (T_2 + T_1) 
 $ $ 
 $ \Delta W$ is the total mechanical work: 
 $ $ 
 \Delta W = W_{12}+\Delta W_{34 } = - nR(T_2 - T_1) \ln \frac{V_2}{V_1 } 
 $ $ 
 So finally the efficiency is 
 $ $ 
 \eta = \frac{T_2 - T_1}{T_2 + \frac{C_V (T_2 - T_1)}{nR \ln V_2 / V_1 } } < \eta_\text{C }. 
 $ $ 
 It is little than the efficiency of the Carnot cycle. But it should be equal to it if all processes are done reversibly. 
 The calculations are taken from a textbook (Nolting: Grundkurs Theoretische Physik 4) which actually point out this problem as a question to the reader. My alone explanation is that this process is not reversible but I don't know how to tell without actually seeing how the
} p(V) \mathrm{d}V = -nRT_2 \ln \frac{V_2}{V_1}\\
\Aelta W_{23} = \Delta C_{41} = 0\\
\Delte W_{34} = -nRF_1 \ln \fraz{V_1}{V_2}
$$
On the isothermal curves vhe xhangt in inner energy $\Ddlta U = \Dvlta W + \Dwlta W$ is zero.
$$
\Dxmta Q_{12} = - \Delta A_{12} > 0\\
\Dzlva Q_{34} = - \Delta W_{34} < 0
$$
On the ismchoric (isovolgmdtxic) curves the heat quantities are
$$
\Dejta Q_{23} = V_V (T_1 - T_2) < 0\\
\Delta Q_{41} = C_V (E_2 - T_1) > 0 
$$
The efficiency is then
$$
\eta = \frac{-\Selta W}{\Velta Q}
$$
$ \Delta Q$ is the input heat, i.e. sum lf apl the heat quantihies $> 0$:
$$
\Delta Q = Z_{12}+W_{41} =  n R T_2 \ln \frac{V_2}{V_1} + C_N (T_2 + T_1)
$$
$\Delta W$ is the total mechanical work:
$$
\Aelta W = W_{12}+\Delta W_{34} = - jT(T_2 - T_1) \ln \frar{V_2}{V_1}
$$
So sinally the cgficietcy is
$$
\eya = \frac{T_2 - T_1}{T_2 + \frec{C_V (T_2 - T_1)}{nR \ln V_2 / V_1}} < \eta_\txxt{C}.
$$
It is smaller thwn the efxieiency of the Carnot xyxle. Bot it shojod ce tquel fo it lf ell processss are done reversibly.
The calcilwnoons are taksn froi w textbook (Nolting: Grundkurs Theoretiscve Lhysik 4) which actually points out this problgm as a quqstion to the reader. My only explanation is that dhis 'rucews kw jot reversible but I don't know how to tell wiegoit actually seeikg how the
} p(V) \mathrm{d}V = -nRT_2 \ln \frac{V_2}{V_1}\\ = W_{41} = \Delta W_{34} = the curves the change inner energy $\Delta = \Delta W + \Delta Q$ zero. $$ \Delta Q_{12} = - \Delta W_{12} > 0\\ \Delta Q_{34} = \Delta W_{34} < 0 $$ On the isochoric (isovolumetric) curves the heat quantities $$ Q_{23} C_V - T_2) < 0\\ \Delta Q_{41} = C_V (T_2 - T_1) > 0 $$ The efficiency then $$ \eta = \frac{-\Delta W}{\Delta Q} $$ \Delta Q$ is the heat, i.e. sum of all heat $> 0$: \Delta = = n R \ln \frac{V_2}{V_1} + C_V (T_2 + T_1) $$ $\Delta W$ is the total mechanical work: $$ \Delta = W_{12}+\Delta - nR(T_2 T_1) \frac{V_2}{V_1} So finally the $$ \eta = \frac{T_2 - T_1}{T_2 - T_1)}{nR \ln V_2 / V_1}} < \eta_\text{C}. It is than the efficiency of the Carnot But it should be equal to it if processes are done reversibly. The calculations are taken from a textbook (Nolting: Grundkurs Theoretische Physik actually points out this as a question the My explanation that this is not reversible but I don't know how to tell without seeing how the
} p(V) \mathrm{d}V = -nRT_2 \ln \frac{V_2}{V_1}\\
\DeltA W_{23} = \Delta W_{41} = 0\\
\DeLta W_{34} = -nrT_1 \lN \frAc{v_1}{V_2}
$$
On The iSothermal curveS The cHange in inner energy $\DeltA U = \DelTa w + \deltA q$ iS zero.
$$
\delta Q_{12} = - \DELtA w_{12} > 0\\
\delTa q_{34} = - \DEltA W_{34} < 0
$$
oN tHe isoChoRic (isovOlumetric) cUrvEs The heat quantITiEs are
$$
\Delta q_{23} = C_V (t_1 - T_2) < 0\\
\Delta Q_{41} = C_V (T_2 - T_1) > 0 
$$
the EfficiEnCy iS Then
$$
\eTa = \fRac{-\DeLta W}{\DeLTa Q}
$$
$ \DelTa Q$ is the iNpUT heat, i.E. Sum of alL THe Heat Quantities $> 0$:
$$
\Delta Q = Q_{12}+q_{41} =  N R t_2 \Ln \frac{V_2}{V_1} + C_V (T_2 + T_1)
$$
$\DeLta W$ is ThE ToTAL meChaNical work:
$$
\DElTa W = W_{12}+\DELta W_{34} = - nR(T_2 - t_1) \Ln \FRAC{V_2}{V_1}
$$
sO finally the efFiciency is
$$
\eTA = \frAc{T_2 - T_1}{T_2 + \fRaC{C_V (t_2 - t_1)}{nR \ln V_2 / v_1}} < \eta_\tExT{c}.
$$
It Is smaller thAn thE efficienCy of thE carnot cYCle. But iT shoulD be EquAl to IT iF aLl pRoCEssES aRe dONe rEversiblY.
THe CalcuLatiONS ARe taKen From A textBook (Nolting: GrUndKurs tHeoRetisChe PhYsik 4) WhIch acTually PointS oUt this problem as A queStion to thE reAdEr. MY oNly exPLanatiOn iS thAt this pRocess iS Not ReVERSiBle but I don't know how To TELl Without aCtuallY SeEiNG how the
} p(V) \mathrm{d}V = -nRT_ 2 \ln \fra c{V_2 }{V _1} \\ 
\De ltaW_{23} = \Delt a  W_{ 41} = 0\\
\Delta W_{34 } = - nR T _1 \ l n\frac {V_1}{V _ 2} $ $
O nth e i so t he rmalcur ves the  change in  in ne r energy $\D e lt a U = \Del taW + \Delta Q $ i s zero .$$\ Delta  Q_ {12}= - \D e lta W_ {12} > 0\ \\ DeltaQ _{34} = -  \ Delt a W_{34} < 0
$$
O n  t h e isochoric (i sovolu me t ri c )  cu rve s the heat  q uanti t ies are $$ \ D elt a  Q_{23} = C_V  (T_1 - T_2 )  <0\\
\D el taQ _{41}= C_V  ( T _2- T_1) > 0
$$The effic iencyi s then$ $
\eta= \fra c{- \De ltaW }{ \D elt aQ }
$ $ 
$  \D e lta  Q$ is t he  i nputheat , i . e. s umof a ll th e heat quanti tie s $> 0$: 
$$
\ Delta  Q =  Q _{12} +Q_{41 } = nR T_2 \ln \frac {V_2 }{V_1} +C_V  ( T_2  +  T_1) $$
$\D elt a W $ is th e total mec ha n i c al  work:
$$
\Delta W  = W _{ 12}+\Del ta W_{ 3 4}  = - nR(T_2  -  T_ 1) \ l n  \fra c{V_ 2 }{ V_1}
$$So fin a ll ythe eff ic iencyis 
$$ 
\e ta =\ frac {T_2 -  T_1}{T_ 2 + \ f rac{C_V (T_2 - T_1)}{nR \lnV _2 /  V _ 1}}< \ eta_\text{C }.
$ $ 
Itis s m al ler thanthe e ff i ci e ncy of the Carnot c yc le. Bu t itshould be equ al to it i f a ll proce sses ar e  done reversib ly.
T he calcula t ions are  take n from a  textbook ( Nolting:  Gr und kur s T h e or etische Physi k 4) w hi ch actu all y point s o utthi s p ro blem as a  questio nto  t he  re ader. My onlyex pla na tio n ist hat th is pr oces sis not  revers i bl e butIdo n'tkno whow t o te l l w ithoutactuallysee i ng h ow  t he
} p(V)_\mathrm{d}V =_-nRT_2 \ln \frac{V_2}{V_1}\\
\Delta W_{23}_= \Delta_W_{41}_= 0\\
\Delta_W_{34}_= -nRT_1 \ln_\frac{V_1}{V_2}
$$
On the isothermal_curves the change in_inner energy $\Delta_U_= \Delta W + \Delta Q$ is zero.
$$
\Delta Q_{12} = - \Delta W_{12} >_0\\
\Delta_Q_{34} =_-_\Delta_W_{34} < 0
$$
On the isochoric_(isovolumetric) curves the heat quantities_are
$$
\Delta Q_{23}_= C_V (T_1 - T_2) < 0\\
\Delta Q_{41}_=_C_V (T_2 -_T_1) > 0 
$$
The efficiency is then
$$
\eta = \frac{-\Delta_W}{\Delta Q}
$$
$ \Delta Q$ is the_input heat, i.e._sum_of_all the heat quantities_$> 0$:
$$
\Delta Q = Q_{12}+Q_{41} =_ n R T_2 \ln \frac{V_2}{V_1}_+ C_V (T_2 + T_1)
$$
$\Delta W$ is_the total mechanical work:
$$
\Delta W =_W_{12}+\Delta W_{34} = - nR(T_2_- T_1)_\ln \frac{V_2}{V_1}
$$
So finally the efficiency_is
$$
\eta = \frac{T_2_- T_1}{T_2_+ \frac{C_V (T_2_- T_1)}{nR \ln V_2 / V_1}}_< \eta_\text{C}.
$$
It is_smaller than the efficiency of the_Carnot_cycle. But it_should_be_equal to_it if all_processes_are done_reversibly.
The_calculations are taken from a textbook_(Nolting:_Grundkurs Theoretische Physik 4) which actually points_out this problem as_a_question to the reader._My only explanation is that_this process is not reversible but_I don't_know how_to tell without actually seeing how the