chemistry, A 8.10-L container holds a mixture of two gases at 51 °C.

Question

A 8.10-L container holds a mixture of two gases at 51 °C. The partial pressures of gas A and gas B, respectively, are 0.308 atm and 0.520 atm. If 0.120 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

PV=nRT

R=0.0821(L*atm)/(K*mol)

T=51+273=324K

V=8.1L

 

Total Pressure equals the sum of the partial pressures of the gasses in the container so initial pressure is sum of the pressure of gas A and gas B.

 

P₁=0.308atm+0.520atm=0.828atmSolve for initial mols of gas
n₁= (0.828*8.1)/(0.0821*324)molAdd 0.120mol of gas to n₁ to get n₂ and use n₂ to solve for P₂ or the total pressure once the third gas has been added.