• # question_answer A vessel of uniform cross-section of length 250 cm as shown in figure is divided in two parts by a weightless and frictionless piston. One part contains 5 moles of He (g) and other part 2 moles of ${{\text{H}}_{\text{2}}}$ (g) and 10 moles of ${{\text{O}}_{\text{2}}}$ (g) added at the same temperature and pressure. In which reaction takes place, finally vessel cooled to 350 K and 1 atm. What is length of compartment? (Assume piston volume and ${{H}_{2}}O(l)$ volume is negligible)                          A)  89.28cm              B)  200 cm              C)  250 cm                 D)  312.5 cm

Idea This question can be solved by using the concept of stoichiometry and volume-mole relationship. ${{H}_{2}}+\frac{1}{2}{{O}_{2}}\xrightarrow[{}]{{}}{{H}_{2}}O$ Since mole of ${{H}_{2}}$ reacts with $\frac{1}{2}$ moles of ${{O}_{2}}$. Hence, net chemical reaction occuring at compartment and moles of ${{O}_{2}}$ remain after reaction is$2{{H}_{2}}+{{100}_{2}}\xrightarrow[{}]{{}}\underset{\begin{smallmatrix} \text{Volume}\,\text{of}\,{{H}_{2}}O \\ \text{will}\,\text{not}\,\text{considered} \end{smallmatrix}}{\mathop{\underset{\downarrow }{\mathop{2{{H}_{2}}O+9{{O}_{2}}}}\,}}\,$ After reaction Moles of He in compartment = 5 Moles of ${{O}_{2}}$ in compartment = 9 Total moles of ${{H}_{2}}$ and ${{O}_{2}}$after reaction = 9+ 5 =14 Since, volume of gases $\propto$ number of moles of gases due to same temperature and pressure. Length of the compartment $=\frac{\text{moles}\,\text{of}\,\text{He}}{\text{Total}\,\text{mole}}\times$ length of full compartment$=\frac{5}{5+9}\times 250$ $=\frac{1250}{14}=89.28\,\text{cm}$ TEST Edge Students are advised to undergo the concept of ideal gas and van der Waal's gas equation also to solve the problem of various kind including conceptual mixing of mole concept and ideal gas which are asked very frequently.