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Provide me with a step by step solution for Question:12

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### Solution to Question 12 (membrane problem)

Chamber I: volume $$V$$, initially N$$_2$$O$$_4$$ at 20 mm Hg, temperature $$T$$. Chamber II: volume $$3V$$, contains liquid water, vapour pressure = 20 mm Hg at $$T$$.
Heated to $$1.2T$$. At $$1.2T$$, 50% of N$$_2$$O$$_4$$ dissociates: N$$_2$$O$$_4$$ ⇌ 2NO$$_2$$.
Initial moles of N$$_2$$O$$_4$$ in I: $$n_0 = \frac{P V}{RT} = \frac{20V}{RT}$$.
After dissociation, degree $$\alpha=0.5$$: total moles = $$n_0 (1 - \alpha) + 2 n_0 \alpha = n_0 (1+\alpha) = 1.5 n_0$$.
Pressure in I if volume fixed: $$P_I' = \frac{n_{\text{total}} R (1.2T)}{V} = 1.5 n_0 \times \frac{R (1.2T)}{V} = 1.5 \times \frac{20V}{RT} \times \frac{1.2RT}{V} = 1.5 \times 20 \times 1.2 = 36$$ mm Hg.
But membrane allows NO$$_2$$ to move to II. At equilibrium, partial pressures of NO$$_2$$ in both chambers equal. Let $$x$$ be the final partial pressure of NO$$_2$$ in both chambers.
In chamber I, from dissociation: N$$_2$$O$$_4$$ pressure = $$P_{N_2O_4}$$, NO$$_2$$ pressure = $$x$$. The total pressure in I = $$P_{N_2O_4} + x$$. Also from dissociation constant: $$K_p = \frac{x^2}{P_{N_2O_4}}$$. But we know initial $$n_0$$ and $$\alpha$$ at $$1.2T$$ before diffusion. However, diffusion changes composition.
Let initial moles N$$_2$$O$$_4$$ = $$n_0$$. After dissociation and diffusion equilibrium, let moles N$$_2$$O$$_4$$ in I = $$n_1$$, moles NO$$_2$$ in I = $$n_2$$, in II = $$n_3$$. Total volume I = $$V$$, II = $$3V$$.
At equilibrium, $$P_{NO_2}$$ same in both: $$\frac{n_2 R (1.2T)}{V} = \frac{n_3 R (1.2T)}{3V}$$ ⇒ $$n_2 = \frac{n_3}{3}$$.
Also total moles of N$$_2$$O$$_4$$ plus NO$$_2$$ conserved: $$n_1 + n_2 + n_3 = 1.5 n_0$$ (since initial dissociation produced 1.5 $$n_0$$ total moles).
Dissociation equilibrium: $$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}}$$. We can find $$K_p$$ from initial condition before diffusion: when $$\alpha=0.5$$, $$P_{N_2O_4} = \frac{(1-\alpha)}{1+\alpha} P_{\text{total}} = \frac{0.5}{1.5} \times 36 = 12$$ mm Hg, $$P_{NO_2} = \frac{2\alpha}{1+\alpha} P_{\text{total}} = \frac{1}{1.5} \times 36 = 24$$ mm Hg. So $$K_p = \frac{24^2}{12} = 48$$ mm Hg.
At diffusion equilibrium, let $$P_{NO_2} = y$$ mm Hg, then $$P_{N_2O_4} = \frac{y^2}{48}$$. In chamber I, total pressure $$P_I = \frac{y^2}{48} + y$$. In chamber II, only NO$$_2$$ and water vapour. Water vapour pressure at $$1.2T$$ = 30 mm Hg. So $$P_{II} = y + 30$$.
Also, total moles of NO$$_2$$: $$n_{NO_2} = n_2 + n_3 = 4 n_2$$ (since $$n_3=3n_2$$). But $$n_2 = \frac{y V}{R (1.2T)}$$. Similarly, moles N$$_2$$O$$_4$$: $$n_1 = \frac{(y^2/48) V}{R (1.2T)}$$.
Total moles = $$n_1 + 4 n_2 = \frac{V}{R (1.2T)} \left( \frac{y^2}{48} + 4y \right)$$. This equals $$1.5 n_0 = 1.5 \times \frac{20V}{RT}$$. Cancel $$\frac{V}{RT}$$: $$\frac{1}{1.2} \left( \frac{y^2}{48} + 4y \right) = 30$$ ⇒ $$\frac{y^2}{48} + 4y = 36$$. Multiply by 48: $$y^2 + 192 y = 1728$$ ⇒ $$y^2 + 192 y - 1728 = 0$$.
Solve: $$y = \frac{-192 \pm \sqrt{192^2 + 4 \times 1728}}{2} = \frac{-192 \pm \sqrt{36864 + 6912}}{2} = \frac{-192 \pm \sqrt{43776}}{2} = \frac{-192 \pm 209.2}{2}$$. Positive root: $$y = \frac{17.2}{2} = 8.6$$ mm Hg.
Then $$P_I = \frac{y^2}{48} + y = \frac{73.96}{48} + 8.6 = 1.54 + 8.6 = 10.14$$ mm Hg. $$P_{II} = y + 30 = 38.6$$ mm Hg.
Thus $$ab = 10$$, $$cd = 39$$ (rounded). So $$abcd = 10 \times 39 = 390$$.
But $$ab$$ and $$cd$$ are two‑digit numbers: $$ab=10$$, $$cd=39$$ ⇒ $$a=1, b=0, c=3, d=9$$, product $$abcd = 1 \times 0 \times 3 \times 9 = 0$$. That seems odd. Possibly $$ab=10$$, $$cd=39$$, then $$abcd = 1039$$? No, $$abcd$$ is product of digits? Likely they mean $$ab$$ and $$cd$$ are two‑digit numbers, and $$abcd$$ is the four‑digit number formed by concatenating them: $$ab=10$$, $$cd=39$$ ⇒ $$abcd=1039$$. Or maybe $$ab=10$$, $$cd=39$$, then $$abcd = 10 \times 39 = 390$$. I think 390.

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Problem #57