A 0.316 mol sample of nitrogen gas, N2 (g), is placed in a 4.00 L container at 315 K.

2.04 atm is the pressure of the nitrogen gas in the container when 0.316 mol sample of nitrogen gas, N2 (g), is placed in a 4.00 L container at 315 K.

Explanation:

Data given:

moles of nitrogen gas, n = 0.316

volume of the nitrogen gas, V = 4 litres

temperature of the container of nitrogen gas  = 315 K

R (gas constant) = 0.08201 Latm/mole K

Pressure of the nitrogen gas on the container,  P = ?

from the data given, we will apply ideal gas law equation to calculate the pressure on the nitrogen gas:

PV = nRT

Rearranging the equation,

P =

P =

P = 2.04 atm

The pressure of the nitrogen gas in the container is 2.04 atm.

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