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?

1 Answer
anor277
Aug 29, 2016

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"#.

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