question_answer

Two gases, each at temperature T, volume V and pressure P are mixed such that the temperature and volume of the mixture are T and V respectively. What would be the pressure of the mixture? Justify your answer on the basis of kinetic theory.

Answer:

                From kinetic theory, \[P=\frac{1}{3}\frac{M}{V}\overline{{{\upsilon }^{2}}}\]                 But  \[\overline{{{\upsilon }^{2}}}\propto T\]      \[\therefore \] \[P\propto \frac{MT}{V}\] As both T and V remain unchanged but mass M is doubled, so the pressure of mixture gets doubled i.e., it is equal to 2P.