Chemistry - Ideal Gas Question with Solution | TestHub
A container is divided into two parts in the ratio 1:3, by a fixed semi-permeable membrane (SPM), which permits movement of NO2 gas only. At T K, compartment - I contains only N2O4 gas at 20 mm Hg pressure and compartment - II contains small amount of liquid water, but sufficient to maintain equilibrium with vapours at any temperature. Now, both the compartments are heated to '1.2T' K at which 50% of N2O4 dissociated into NO2. The vapour pressure of water at T K and '1.2T' K are 20 and 30 mm Hg, respectively. If ab and cd are the final pressures (in mm Hg) of gases in compartments I and II, respectively, then the value of abcd is
Answer:
Solution:
Let compartment volumes be V (I) and 3V (II). Initial moles of N2O4 in I: n0 = 20V/RT (mm-L units). On heating to 1.2T with 50% dissociation (N2O4 → 2 NO2): N2O4 left = 0.5 n0, staying in compartment I (membrane blocks it): p(N2O4) = 0.5 × 20 × 1.2 = 12 mm Hg. NO2 formed = n0 moles; it distributes over the whole volume V + 3V = 4V at 1.2T with equal partial pressure on both sides: p(NO2) = n0 R(1.2T)/(4V) = 20 × 1.2/4 = 6 mm Hg. Compartment I total pressure = 12 + 6 = 18 mm Hg, so ab = 18. Compartment II holds saturated water vapour (liquid present) at 30 mm Hg plus NO2 at 6 mm Hg: total = 30 + 6 = 36 mm Hg, so cd = 36. Hence abcd = 1836.
