A 7.20 L container holds a mixture of two gases at 29 °C. The partial pressures of gas A and gas B, respectively, are 0.179 atm and 0.749 atm. If 0.200 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

P (total) = 1.63 atm

Explanation:

Given dа ta:

Volume of container = 7.20 L

Temperature = 29°C

Number of moles of third gas = 0.200 mol

Pressure of gas A = 0.179 atm

Pressure of gas B = 0.749 atm

Total pressure = ?

Solution:

According to the Dalton's law of partial pressure,

"The total pressure exerted by the mixture of gases is the sum of partial pressure of the individual gases"

P (total) = P₁ + P₂ + P₃ + ... Pₙ

First of all we will determine the pressure of third gas.

PV = nRT

P = nRT / V

P = 0.200 mol * 0.0821 atm. L. mol⁻¹. k⁻¹ * 302 K / 7.20 L

P = 5 atm. L / 7.20 L

P = 0.7 atm

P (total) = PA + PB + P₃

P (total) = 0.179 + 0.749 + 0.7

P (total) = 1.63 atm