A 8.00-L container holds a mixture of two gases at 23 °C. The partial pressures of gas A and gas B, respectively, are 0.349 atm and 0.819 atm. If 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
Calculate mols gas A from PV = nRT. You know its partial pressure, volume, R, T (remember T must be in kelvin).
Calculate mols gas B from PV = nRT in a similar manner.
Add mols gas A, mols gas B + 0.210 from the third gas. Use PV = nRT using total mols and solve for P which will be total P.
Post your work if you get stuck.
To find the total pressure after adding the third gas, we can use Dalton's law of partial pressures, which states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases.
In this case, the initial partial pressures of gas A and gas B are given as 0.349 atm and 0.819 atm, respectively. To find the total pressure, we need to calculate the sum of these partial pressures.
Partial pressure of gas A + Partial pressure of gas B = Total pressure
0.349 atm + 0.819 atm = Total pressure
Adding the partial pressures, we get:
0.349 atm + 0.819 atm = 1.168 atm
So, initially, the total pressure in the container is 1.168 atm.
Now, 0.210 mol of a third gas is added to the container. Since there is no change in volume or temperature, the total pressure will not change. Therefore, the total pressure after adding the third gas will also be 1.168 atm.