Question

A 7.65-L container holds a mixture of two gases at 25 °C. The partial pressures of gas A and gas B, respectively, are 0.189 atm and 0.696 atm. If 0.210 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

initial partial pressure A=0.189 atm , B=0.696 atm

total pressure= 0.189+ 0.696 =0.885 atm

PV=nRT

P= 0.885 atm , V= 7.65 L, R=0.082057 L atm mol^-1 K^-1, n=? ,T=298 K

n= PV/RT= (0.885 atm * 7.65 L) / (0.082057 L atm mol^-1 K-1 * 298 K )= 0.2768 mol

after this, 0.210 mol of gas C is added to it.

total moles of gas= 0.2768 + 0.210 =0.4868 mol

new pressure P =nRT/V= (0.4868 mol * 0.082057 L atm mol^-1 K-1 * 298 K ) / (7.65 L) = 1.556 atm

ans : pressure of the system is 1.556 atm.