Answer to Question #263542 in General Chemistry for diamond

Question

7.30 L container holds a mixture of two gases at 43 °C.

43 °C. The partial pressures of gas A and gas B, respectively, are 0.272 atm

0.272 atm and 0.859 atm.

0.859 atm. If 0.210 mol

0.210 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=7.30L

 

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 moles of gas

n₁= (0.828*8.1)/(0.0821*324)mol Add 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