Chemistry Labs

Problem 2

Dissociating gas cycle. Dinitrogen tetroxide forms an equilibrium mixture with nitrogen dioxide: NX2OX4(g)⇌2 NOX2(g)\ce{N2O4(g) <=> 2 NO2(g)}. 1.00 mol of NX2OX4\ce{N2O4} was placed in an empty vessel of fixed volume 24.44 dm3^3. The equilibrium gas pressure at 298 K was 1.190 bar; when heated to 348 K the equilibrium pressure rose to 1.886 bar. (a) Calculate ΔG∘\Delta G^{\circ} of the reaction at 298 K, assuming ideal gases. (b) Calculate ΔH∘\Delta H^{\circ} and ΔS∘\Delta S^{\circ}, assuming they do not vary significantly with temperature. The reversible dissociation of NX2OX4\ce{N2O4} can be exploited in power cycles: in step 3→43 \to 4 the hot gas mixture expands reversibly and adiabatically through a turbine. (c) Give the equation for the work done by 1 mol of an inert gas (air) during the reversible adiabatic expansion 3→43 \to 4, assuming constant CV,mC_{V,m} and a temperature drop from T3T_3 to T4T_4.
Step 5 of 5: Work of adiabatic expansion
wair=ΔU=CV,m(air) (T3−T4)w_{\text{air}} = \Delta U = C_{V,m}(\text{air})\,(T_3 - T_4)
Analysis

For a reversible adiabatic expansion q=0q = 0, so the work done by the gas equals its internal-energy decrease: wair=−ΔU=CV,m(T3−T4)w_{\text{air}} = -\Delta U = C_{V,m}(T_3 - T_4) (taking work output as positive).

Common pitfall. Do not forget that pinip_{ini} must be recalculated at each temperature with the ideal-gas law — using 1.014 bar at 348 K gives the wrong xx.