Question

A 9.70 L container holds a mixture of two gases at 53C. The partial pressures of gas A and gas B, respectively, are 0.368 atm and 0.893 atm. If 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

This is an illustration of `"Dalton's Law of Partial Pressures......"`

Explanation:

`"Dalton's Law of Partial Pressures......"`, `"which states, that in a"` `"gaseous mixture, the partial pressure of a gaseous component is"` `"the same as the pressure it would exert if it alone occupied"` `"the container."`

And thus `P_"mixture"=P_A+P_B+P_c`

We are given `P_A`, and `P_B`, but we must work out `P_C`, which we can do by assuming ideality and employing the equation:

`P_C=(0.210*molxx0.0821*(L*atm)/(K*mol)xx326*K)/(9.70*L)=0.579*atm.`

And so.....`P_"mixture"=P_A+P_B+P_c`

`=(0.368+0.893+0.579)*atm`

`=1.82*atm.`