Question

The stopcock connecting a 2.14 L bulb containing oxygen gas at a pressure of 8.19 atm, and a 9.84 L bulb containing krypton gas at a pressure of 2.65 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is atm.
what is the final pressure of the system in atm?

Answer 1

Answer : The final pressure of the system in atm is, 3.64 atm

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$

or,

$P_1V_1+P_2V_2=P_fV_f$

where,

$P_1$ = first pressure = 8.19 atm

$P_2$ = second pressure = 2.65 atm

$V_1$ = first volume = 2.14 L

$V_2$ = second volume = 9.84 L

$P_f$ = final pressure = ?

$V_f$ = final volume = 2.14 L  + 9.84 L = 11.98 L

Now put all the given values in the above equation, we get:

$8.19atm\times 2.14L+2.65atm\times 9.84L=P_f\times 11.98L$

$P_f=3.64atm$

Therefore, the final pressure of the system in atm is, 3.64 atm