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:

$(9.61atm)\times (3.06L)=n_1\times RT$

$n_1=\frac{29.4}{RT}$

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:

$(1.75atm)\times (6.65L)=n_2\times RT$

$n_2=\frac{11.6}{RT}$

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

$P_{total}=(n_1+n_2)\times \frac{RT}{V_{total}}$

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:

$P_{total}=(\frac{29.4}{RT}+\frac{11.6}{RT})\times \frac{RT}{9.71L}$

$P_{total}=4.22atm$

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