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?

1 Answer
May 9, 2017

Answer:

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.#