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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases

Gas-A is kept in the left partition of the container with a volume V/n at pressure P. Gas-B is kept in the right part of the container with a volume of V/(n+1) at pressure P. Both the portions are the same temperature T. When what is the value of the Pressure P' when the partition is removed?

(A) P (B) nP (C) n(n+1)P (D) 2P

1 Answer

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For gas A, $n_A = \large\frac{P \times V}{nRT}$ and $n_B = \large\frac{P \times V}{(n+1)RT}$
After oprning the partition, $n_A + n_B = n = \large\frac{P' \times (V/n+V/(n+1))}{RT}$
$\Rightarrow P' = P$
answered Jan 31, 2014 by balaji
edited Jan 31, 2014 by balaji

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