Thermodynamics

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

T = 300 K

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

mT = 8.3 J/K1 mol1

Now, using Ideal gas equation

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

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

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

$P_1=2*10^{7}Pa$

$P_1{v}_1=P_2{v}_2$

Since v₂ = 2v₁ Hence   $P_2=\frac{P_1}{2}\left(IsothermalExpansion\right)$

$P_2=1*10^{7}Pa$

$P_2{\left(v_2\right)}^y=P_3{\left(2v_2\right)}^y$

$P_3=\frac{1*10^7}{2^{1.5}}=3.536*10^{6}Pa$

Efficiency $\eta =50\mathrm{\%}$ $$

$\Rightarrow \eta =\left(1-\frac{T_2}{T_1}\right)*100=50$

$\frac{{}_{}}{}$ $\Rightarrow 1-\frac{T_2}{T_1}=\frac{50}{100}\Rightarrow 1-\frac{T_2}{T_1}=\frac{1}{2}$  -(i)

After the change :- $1-\frac{T_2^1}{T_1}=\frac{50*\left(1.3\right)}{100}=\frac{65}{100}=.65$ $\frac{{}_{}^{}}{{}_{}}\frac{}{}\frac{}{}$ -(ii)

$\Rightarrow 1-\left(\frac{T_2-40}{T_1}\right)=.65$ --(ii)

$\Rightarrow \frac{40}{T_1}=.65-.50=.15\Rightarrow T_1=\frac{4000}{15}=266.7k$

$\mathrm{n}=\frac{\mathrm{W}}{{\mathrm{Q}}_{\mathrm{h}}}=1-\frac{300}{900}=\frac{2}{3}$

${\mathrm{Q}}_{\mathrm{h}}=\frac{3}{2}\mathrm{W}=1800\mathrm{J}$

${\mathrm{Q}}_{\mathrm{L}}={\mathrm{O}}_{\mathrm{h}}-\mathrm{W}=600\mathrm{J}$

Let x gm is burnt

Moles = x/280

Heat released by x/280 mol = 2.5 * 0.45 kJ

Heat released by 1 mol =  $\frac{2.5*0.45*280}{x}=kJ$

$\begin{array}{l}\Delta H=\Delta U+\Delta ngRT\\ \Delta H=\Delta U\end{array}$            

$9=\frac{2.5*280*0.45}{x}$            

x = 35 gm

3C (Graphite) + 4H₂ (g) -> C₃H₈ (g)

$\Delta H_f\left(C_3{H}_8\right)=\left[3*\Delta H_{comb}\left(C\right)\right]+\left[4*\Delta H_{comb}\left(H_2\right)\right]-\left[\Delta H_{comb}\left(C_3{H}_8\right)\right]$        

= -10.3.7 kJ/mole

For adiabatic process – PV^Y = const

$T_1{V_1}^{Y-1}=T_2{V}_2^{Y-1}$

$\frac{T_2}{T_1}={\left(\frac{v_1}{v_2}\right)}^{Y-1}={\left(\frac{d_2}{d_1}\right)}^{Y-1}={\left(32\right)}^{\left(\frac{7}{5}-1\right)}={\left(32\right)}^{2/5}$

$={\left(2\right)}^2=4$

2O₃(g) -> 3O₂

t = 0      a moles               0

-0.5a mole     +0.75a mole

At Eq.    0.5a mole         0.75a mole

Total moles at eq = 0.5a + 0.75 a = 1.25a

$P_{O_3}=x_{O_3}*P_T=\frac{0.5a}{1.25a}*1=\frac{2}{5}atm$

$P_{O_2}=x_{O_2}*p_T=\frac{0.75a}{1.25a}*1=\frac{3}{5}atm$

$K_p=\frac{{\left(P_{O_2}\right)}_{eq}^3}{{\left(P_{O_3}\right)}_{eq}^3}=\frac{{\left(\frac{3}{5}\right)}^3}{{\left(2/5\right)}^2}=1.35atm$

$\Delta G^{°}=-RT\mathcal{l}nKP$

$\begin{array}{l}=-8.3*300\mathcal{l}n1.35\\ =-747J/mole\end{array}$

T₁ = 727 + 273 = 1000k

T₂ = 127° + 273 = 400 k

Q₁ = 5 * 10³ k cal

$\Rightarrow \frac{600}{1000}=\frac{w}{5000}$                

w = 12.6 * 10⁶ J