Thermodynamics

Using Ideal gas Equation :

PV = nRT

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

=> $m_1=0.12,m_2=0.64$

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

Millimoles of HCl = 200 × 0.2 = 40

Millimoles of NaOH = 300 × 0.1 = 30

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

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

Mass of solution = 500 × 1 g

= 500 g

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

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

$=\frac{1713}{500\times 4.18}°C$  

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

Ans. = 82

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

= 10 J

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

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

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

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

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

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

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

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

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

$=-632kJ/mole$

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 =?

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

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

$\Rightarrow Q_1=16\times 10^{6}J$