A7.20 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 data:

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

the atomic mass is the bottom number on the element.

-steel jelly

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole