Question

Suppose that 1.28 g of neon gas and 2.49 g of argon gas are confined in a 9.87-L container at 27 C. What would be the pressure in the container?

Answer 1

Approx. `0.3*atm`

Explanation:

We use `"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; the total pressure is the SUM of the individual partial pressures.

And thus `P_"Ne"` `=` `(n_"Ne"RT)/V` `=` `((1.28*g)/(20.18*g*mol^-1)xx0.0821*L*atm*K^-1*mol^-1xx300*K)/(9.87*L)` `=` `0.158*atm`

And `P_"Ar"` `=` `(n_"Ar"RT)/V` `=` `((2.49*g)/(39.95*g*mol^-1)xx0.0821*L*atm*K^-1*mol^-1xx300*K)/(9.87*L)` `=` `0.156*atm`

`P_"Total"` `=` `P_"Ne"+P_"Ar"` `=` `??*atm`

Of course, I could have simply worked out the molar quantity of the combined gases, and solved the problem directly. This treatment illustrates `"Dalton's Law"`.