Thermodynamics

From   $\Delta H=\Delta G+T\Delta S\therefore \Delta S=\frac{\Delta H-\Delta G}{T}$

$\therefore$ $\Delta S=\left[\frac{51.4-\left(-49.4\right)}{300}\right]*1000J{K}^{-1}mo{l}^{-1}=336J{K}^{-1}mo{l}^{-1}$

Since process is isothermal, so

$W=nRT\mathcal{l}n\left(\frac{V_f}{V_i}\right)=1*8.3*300*\mathcal{l}n\left(\frac{4}{2}\right)=1*8.3*300*\mathcal{l}n\left(2\right)$       

$\Rightarrow W=1*8.3*300*0.6931\approx 1725.82J\approx 17258*10^{-1}J$

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