Suppose you are collecting a gas over water at 23 oC and obtain a pressure reading of 785.8 mm Hg. If the vapor pressure of water is 21.1 mm Hg at the collection temperature, what is the pressure of the collected gas

764.7 mmHg

Explanation:

Step 1: Given data

Step 2: Calculate the partial pressure of the gas collected

The total pressure of the gaseous mixture is equal to the sum of the gas collected and the water vapor.

P = pH₂O + pGas

pGas = P - pH₂O

pGas = 785.8 mmHg - 21.1 mmHg = 764.7 mmHg

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