An ideal gas present in a container connected to frictionless, weightless pistion is present at a fixed pressure P . The expected graph for the variation of volume with temperature was to be like as i

Question

An ideal gas present in a container connected to frictionless, weightless pistion is present at a fixed pressure P . The expected graph for the variation of volume with temperature was to be like as in (i), however actual observations revealed that the graph between V & T to be like (ii). The deviation was attributed to dissociation of gas as $D_{m} (g) ⇌ mD (g)$ . If it is known that percentage dissociation of $D_{m} (g)$ is $60%$ . Find the value of m .

TestHub · Accepted Answer

For actual D m ​ ( g ) ⇌ m D ​ ( g ) t = 0 , 1 0 t = t e q ​ . ( 1 − 0.6 ) m ( 0.6 ) = 0.4 n t o t ​ = 0.4 + m ( 0.6 ) ...(1) Slope of the curve d T d V ​ = P n R ​ . 20 49.1 − 47.9 ​ = P ( 0.4 + m ( 0.6 )) R ​ .. ( 2 ) For the expected graph, slope of curve : \displaylines 20 50.2 − 49.9 ​ = P R ​ 20 0.3 ​ = P R ​ .. ( 3 ) From equations (2) and (3): 20 1.2 ​ = ( 0.4 + m ( 0.6 )) 20 2.3 ​ ⟹ m = 6

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