A 8.65-L container holds a mixture of two gases at 11 °C. The partial pressures of gas A and gas B, respectively, are 0.205 atm and 0.658 atm. If 0.200 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

The total pressure = 1.402 atm

calculation

Total pressure = partial pressure of gas A + partial pressure of gas B + partial pressure of third gas

partial pressure of gas A = 0.205 atm

Partial pressure of gas B = 0.658 atm

partial pressure for third gas is calculated using ideal gas equation

that is PV=nRT where,

p (pressure) = ? atm

V (volume) = 8.65 L

n (moles) = 0.200 moles

R (gas constant) = 0.0821 L. atm/mol. k

T (temperature) = 11°c into kelvin = 11+273 = 284 k

make p the subject of the formula by diving both side by V

p = nRT/v

p = [ (0.200 moles x 0.0821 L. atm/mol. K x 284 K) / 8.65L) ] = 0.539 atm

Total pressure is therefore = 0.205 atm + 0.658 atm + 0.539 atm

=1.402 atm