A 7.35-L container holds a mixture of two gases at 17 °C. The partial pressures of gas A and gas B, respectively, are .415 atm and .720 atm. If .160 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

1 Answer
Dec 4, 2015

Answer:

Dalton's law of partial pressures states that in a gaseous mixture, the pressure exerted by a component is the same as the pressure it would exert if it alone occupied the container.

Explanation:

Since pressures are additive, and we know #P_A# and #P_B#, all we need to do is calculate #P_C#.

#P_C = (n_CRT)/V# #=# #(0.160cancel(*mol)xx0.0821cancel(*L)(*atm)*cancel(K^-1)*cancel(mol^-1)xx300cancel(K))/(7.35cancel(L))# #=# #??# #atm#

#P_(TOTAL) = P_C + P_A + P_B=??#