Question

Consider two bulbs seperated by a valce. Both bulbs are amintained at the same temperature. Assume that when the valve between the two bulbs is closed, the gases are sealed in their respective bulbs. When the valve is closed, the following data apply:
Bulb A Bulb B
Gas Ne CO
V 2.50L 2.00L
P 1.09 atm 0.73 atm


Assuming no temperature change, determine the final pressure inside the system after the valve connecting the two bulbs is opened. Ignore the volume of the tube connecting the two bulbs.

Answer 1

Answer:

The  pressure is $P_f = 0.93 \ atm$

Explanation:

From the question we are told that

   The  volume of  Ne  is  $V_N = 2.50 \ L$

    The volume  of  CO is  $V_C = 2.00 \ L$

    The  pressure of  $Ne$ is  $P_N = 1.09 \ atm$

      The  pressure of  CO is $P_C = 0.773 \ atm$

The  number of  moles of  Ne present is evaluated using the ideal gas equation as

      $n_N = \frac{P_N * V_N}{R T}$

=>   $n_N = \frac{1.09 * 2.50 }{R T} = \frac{2.725}{RT}$

The  number of  moles of  CO present is evaluated using the ideal gas equation as

      $n_N = \frac{P_C * V_C}{R T}$

=>   $n_N = \frac{0.73 * 2.00 }{R T} = \frac{1.46}{RT}$

The  total number of moles of gas present is evaluated as

        $n_T = n_N + n_C$

        $n_T = \frac{2.725}{RT} + \frac{1.46}{RT}$

      $n_T = \frac{4.185}{RT}$

The  total volume of gas present when valve is opened is  mathematically represented as

                $V_T = V_N + V_C$

    =>        $V_T = 2.50 + 2.00 = 4.50 \ L$

So

  From the ideal gas equation the final pressure inside the system  is mathematically represented as

         $P_f = \frac{n_T * RT }{ V_T}$

=>      $P_f = \frac{[\frac{4.185}{RT} ] * RT }{ 4.50}$

=>       $P_f = 0.93 \ atm$