When a sample of a gas was placed in a sealed container with a volume of 3.35L and heated to 105°C, the gas vaporized and the resulting pressure inside the container was 170.0kPa. How many moles of the gas was president?

181.12 moles

Explanation:

V = 3.35L

T = 105°C = (105 + 273.15)K = 378.15K

P = 170.0kPa = 170.0*10³Pa (1kPa = 1000Pa)

R = 8.314J /mol.K

n = ?

From ideal gas equation,

PV = nRT

P = pressure of the gas

V = volume of the gas

n = number of moles

R = ideal gas constant

T = temperature of the gas

PV = nRT

n = PV / RT

n = (170*10³ * 3.35) / (8.314 * 378.15)

n = 569500 / 3144.317

n = 181.12 moles

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