Question

The stopcock connecting a 3.06 L bulb containing methane gas at a pressure of 9.61 atm, and a 6.65 L bulb containing oxygen gas at a pressure of 1.75 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is atm.

Answer 1

Answer : The final pressure in the system is 4.22 atm.

Explanation :

First we have to calculate the moles of methane.

`PV=n_1RT`

where,

P = pressure of gas = 9.61 atm

V = volume of gas = 3.06 L

T = temperature of gas = T

`n_1` = number of moles of methane gas = ?

R = gas constant

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

`(9.61atm)* (3.06L)=n_1* RT`

`n_1=(29.4)/(RT)`

Now we have to calculate the moles of oxygen gas.

`PV=n_2RT`

where,

P = pressure of gas = 1.75 atm

V = volume of gas = 6.65 L

T = temperature of gas = T

`n_2` = number of moles of oxygen gas = ?

R = gas constant

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

`(1.75atm)* (6.65L)=n_2* RT`

`n_2=(11.6)/(RT)`

Now we have to determine the final pressure in the system after mixing the gases.

`P_(total)=(n_1+n_2)* (RT)/(V_(total))`

where,

`P_(total)` = final pressure of gas = ?

`V_(total)` = final volume of gas = (3.06 + 6.65)L = 9.71 L

T = temperature of gas = T

R = gas constant

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

`P_(total)=((29.4)/(RT)+(11.6)/(RT))* (RT)/(9.71L)`

`P_(total)=4.22atm`

Therefore, the final pressure in the system is 4.22 atm.