Thermodynamics

η = 1 - T? /T?
1/6 = 1 - T? /T? (i)
2η = 1/3 = 1 - (T? -62)/T? (ii)
By (i) and (ii)
(T? /T? ) = 5/6
1/3 = 1 - (T? /T? ) + 62/T? = 1 - 5/6 + 62/T?
1/3 = 1/6 + 62/T?
1/6 = 62/T?
T? = 62 * 6 = 372K = 99°C

  $W_{AB}=nRTln\frac{2V_1}{V_1}=nRTln2.$       

  $W_{BC}=P_2\left(V_1-2V_1\right)=-P_2{V}_1=-\frac{1}{2}nRT$          

[At B, 2P₂ V₁ = nRT]

$W_{CA}=0$           [CA is isochoric process].

$W_{ABCA}=W_{AB}+W_{BC}+W_{CA}=nRT\left(\mathcal{l}n\left(2\right)-\frac{1}{2}\right)$            

$\Delta Q=\Delta U+\Delta W$

^{$\Rightarrow Q=\Delta U+\frac{Q}{5}\Rightarrow \Delta U=\frac{4Q}{5}=n{C}_v\Delta T\Rightarrow \frac{4Q}{5}=\frac{5R}{2}\Delta T\Rightarrow \Delta T=\frac{8Q}{25R}$}

^{$Q=nc\Delta T=1*C*\frac{80}{25R}\Rightarrow C=\frac{25R}{8}\Rightarrow x=25$}

$V=k{T}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

$T{V}^{-\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=K$

$\Rightarrow \gamma -1=-3/2$

$\Rightarrow \gamma =-1/2$

Work done = $\frac{nR\Delta T}{-y+1}=\frac{1*R*90}{3/2}=6R$

n = 60

U = 3PV + 4

$\Rightarrow n{C}_{v}T=3PV+4$

$\Rightarrow n*\frac{fRT}{2}=3PV+4$

$\Rightarrow f=6+\frac{8}{PV}$

$\Rightarrow f>6\Rightarrow polyatomic$

n = 1 mole

$W_{AB}=nRTln\frac{v_2}{v_1}=1*RTln\frac{2v_1}{v_1}=RTln2$

$W_{BC}=0$

$W_{CA}=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{\frac{P_1}{4}*2V_1-P_1*V_1}{\gamma }$

$\Rightarrow W_{CA}=\frac{-RT}{2\left(\gamma -1\right)}$

$\therefore W_{total}=RT\left[\mathcal{l}n2-\frac{1}{2\left(\gamma -1\right)}\right]$

P = kV³ Þ PV^-3 = k Þ Polytropic Process with index m = -3

W = $\frac{nR\left(T_2-T_1\right)}{1-m}=\frac{nR\left(300-100\right)}{1-\left(-3\right)}=50nR$ .

At constant pressure,

$dU=n{C}_{V}dT=n.\frac{5}{2}RdT$

$dQ=n{C}_{P}dT=n.\frac{7}{2}RdT$

dW = nRdT

dU : dQ : dW = 5 : 7 : 2

Since process is isochoric

So    $\Delta U=n{C}_v\Delta T$

$\Delta U=n\left(\frac{5}{2}R\right)\Delta T--(i)\left[C_V=\frac{5}{2}R\right]$      

And external work

$\Delta W=nR\Delta T--\left(ii\right)$    

$\frac{5}{2}=\frac{x}{10}\Rightarrow x=25.00$                

from newtan's law of cooling :

$\frac{dT}{dt}=-k\left(T-T_0\right)$

Using average value :

$\frac{75-65}{5}=-k\left[\frac{\left(75+65\right)2}{2}-25\right]$

$\frac{\left(65-T\right)}{5}=-k\left[\frac{\left(65+T\right)}{2}-25\right]$

$\Rightarrow \frac{10}{65-T}=\frac{45}{\frac{65+T}{2}-25}$

$\Rightarrow T=57°C$