Thermodynamics

(a) $\Delta G=\Delta G°+RT\mathcal{l}nQ$

$\Delta G°=\Delta H°-T\Delta S°..............\left(1\right)$                      

At equilibrium $\Delta G=0,Q=K_{eq}$

  $\therefore \Delta G°=-RT\mathcal{l}n{K}_{eq}.................\left(2\right)$                   

$\therefore \Delta H°-T\Delta S°=-RT\mathcal{l}n{K}_{eq}$                      

$\mathcal{l}nKeq=-\left(\frac{\Delta H°-T\Delta S°}{RT}\right)$         

(b) $W_{rev}=-nRT\mathcal{l}n\left(\frac{V_f}{V_i}\right)$  

(c) $\Delta G=-T\Delta S_{Total}$ (at constant P)

$\therefore \frac{\Delta G}{\Delta S_{Total}}=-T$        

(d) $\Delta G°=-RT\mathcal{l}nKeq$

$\therefore Keq=e^{\left(-\Delta G°/RT\right)}$  

Adiabatic equation : $P_1{V}^{\gamma }=P_2{V}^{\gamma }\Rightarrow \left(200\right){\left(1200\right)}^{\gamma }=P{\left(300\right)}^{\gamma }$

$\Rightarrow P=\left(200\right){\left(4\right)}^{\frac{3}{2}}=1600KPa$

$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{240-480}{1.5-1}=-480J$  

Using Ideal gas Equation :

PV = nRT

$\Rightarrow n=\frac{PV}{RT}=\frac{400*10^3*500*10^{-6}}{8.3*100}=0.008$    

=> $m_1=0.12,m_2=0.64$

=> $\frac{m_2}{m_1}=\frac{16}{3}$

Millimoles of HCl = 200 * 0.2 = 40

Millimoles of NaOH = 300 * 0.1 = 30

Heat released =  $\frac{30}{1000}*57.1*1000J=1713J$

$\left[\begin{array}{l}\rho =\frac{m}{v}\\ m=\rho v\end{array}\right]$

Mass of solution = 500 * 1 g

= 500 g

Specific heat of water = 4.18 Jg^-1 K^-1

$\Delta T=\frac{q}{mc}$

$=\frac{1713}{500*4.18}°C$  

$=81.96*10^{-2}°C\approx 82*10^{-2}°C$

Ans. = 82

$\frac{\Delta U}{t}=\frac{6000}{60}-90$

= 10 J

$t=\frac{\Delta U}{10}=\frac{2.5*10^3}{10}=2.5*10^2$ second

$\frac{Q_H}{Q_L}=\frac{T_H}{T_L}$

$\Rightarrow Q_L=\frac{T_L}{T_H-T_L}*\left(Q_H-Q_L\right)$

$\Rightarrow \frac{d{Q}_L}{dt}=\frac{T_L}{T_H-T_L}.\frac{d\left(Q_H-Q_L\right)}{dt}$

$=\frac{273-10}{25-\left(-10\right)}*35$

$=263J{s}^{-1}$

$\Delta H=3{\left(\Delta H_{comb}\right)}_{C_2{H}_2\left(g\right)}-{\left(\Delta H_{comb}\right)}_{C_6{H}_6\left(l\right)}$

$=-1300*3-\left(-3268\right)kJ/mole$

$=\left(-3900+3268\right)kJ/mole$

$=-632kJ/mole$

For spontaneous process   $\Delta G<0$

For Isobaric process;    $\Delta P=0$

For Isothermal process;    $\Delta T=0$

$\Delta H$ reaction =  ${\displaystyle \sum _{}^{}B.E_{\left(R\right)}.-{{\displaystyle \sum _{}^{}B.E.}}_{\left(P\right)}}$

T₂ = 200 k

$T_1=527°C+273=800K$      

w = 12000 kJ = 12 * 10⁶ J

Q₁ =?

  $x=1-\frac{T_2}{T_1}=\frac{w}{Q_1}=1-\frac{200}{800}=\frac{12*10^6}{Q_1}$              

$\Rightarrow \frac{3}{4}=\frac{12*10^6}{Q_1}$      

$\Rightarrow Q_1=16*10^{6}J$        

$V_T=2000c{m}^3=2000*10^{-6}{m}^3\left(Totalvolumeofmixture\right)$

T = 300 K

${}_{}{}^{}{}^{}$  $P_T=100kPa=100*10^{3}Pa=10^5$ Pa

mT = 8.3 J/K1 mol1

Now, using Ideal gas equation

$P_{T}V=\left(x_1+x_2\right)RT$

$\Rightarrow 10^5*200*10^{-6}=\left(x_1+x_2\right)*8.3*300$

$\frac{x_1}{x_2}=\frac{0.06}{0.02}=3:1$