Answer to Question #95403 in General Chemistry for Alexis

Question

A student is in the lab performing an experiment using a 7.45 L container that holds a mixture of two gases at 13.0 °C. The partial pressures of gas A and gas B, respectively, are 0.363 atm and 0.834 atm. The student then adds a 0.120 mol quantity of a third gas to the mixture. There is no change in volume or temperature. The student calculates the partial pressure of gas C using the ideal gas law. The student's calculation is shown below. What has the total pressure of the gases become?

Answer 1

$P_{total}=P_A+P_B=1.197\ atm$

Now from ideal gas equation

$PV=nRT\\n=\frac{PV}{RT}=\frac{1.197\times 7.45}{0.082057\times 286}=0.38 \ mole$

so,

$n_A+n_B=0.38$

Now ,

$P°_C=\frac{n_CRT}{V}=\frac{0.12\times 0.0821\times 286}{7.45}=0.3782\ atm\\P_C= x_C\times P°_C=\frac{n_C}{n_A+n_B+n_C}\times 0.3782$

$=\frac{0.12}{0.38+0.12}\times 0.3782=0.09\ atm$

Partial fraction of gas C = $0.09\ atm$