An ideal gas, , is taken in a container connected to frictionless, massless piston. The pressure of — States of Matter and Gaseous State Chemistry Question
Question
An ideal gas, $X_n$, is taken in a container connected to frictionless, massless piston. The pressure of gas is constant (always equal to the atmospheric pressure). The expected graph for the variation of volume of gas with the change in temperature was to be like fig. I (V goes from 49.9 L to 50.2 L over 20°C), but the actual graph is like fig II (V goes from 47.9 L to 49.1 L over 20°C). The deviation was attributed to dissociation of gas as: $X_n(g) \rightarrow n X(g)$. If the yield of this reaction is only 60%, the value of n is
💡 Solution & Explanation
Slope of V vs T is proportional to total moles. Expected slope $m_{exp} = \frac{50.2 - 49.9}{20} = 0.015$. Actual slope $m_{act} = \frac{49.1 - 47.9}{20} = 0.060$. Ratio of actual moles to expected moles is $0.060 / 0.015 = 4$. For reaction $X_n \rightarrow nX$ at 60% yield, 1 mole becomes $(1 - 0.6) + 0.6n = 0.4 + 0.6n$ moles. Thus, $0.4 + 0.6n = 4 \implies 0.6n = 3.6 \implies n = 6$. Therefore, correct answer is 6.