Question

A 9.75 L 9.75 L container holds a mixture of two gases at 41 ° C. 41 °C. The partial pressures of gas A and gas B, respectively, are 0.419 atm 0.419 atm and 0.589 atm. 0.589 atm. If 0.240 mol 0.240 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

The total pressure P = 1.642 atm

Step-by-step explanation:

From the dalton's law

Total pressure of the mixture is

`P = P_(A) + P_(B) + P_(C)` ----- (1)

`P_A = 0.419` atm

`P_B = 0.589` atm

`P_C = (nRT)/(V)`

n = 0.24 mole

R =
`0.0821`
`(L.Atm)/(K.mol)`

T = 41 °c = 314 K

`P_C = ((0.24)(0.0821)(314))/(9.75)`

`P_C` = 0.634 atm

From equation (1)

P = 0.419 + 0.589 + 0.634

P = 1.642 atm

Thus the total pressure P = 1.642 atm