6) At constant temperature, the pressure of a gas is 780 mmHg and
volume of the gas is 2.0 L. Calculate the volume of the gas when
pressure decreases to 620 mmHg.

2.52 L

Explanation:

Step 1:

The following data were obtained from the question:

Initial pressure (P1) = 780 mmHg

Initial volume (V1) = 2.0 L.

Final pressure (P2) = 620 mmHg.

Final volume (V2) =?

Step 2:

Determination of the final volume of the gas.

From the question given, we discovered that the temperature is constant. Since the temperature is constant, the gas is simply obeying Boyle's law.

Applying the Boyle's law equation, the final volume of the gas can be obtained as follow:

P1V1 = P2V2

780 x 2 = 620 x V2

Divide both side by 620

V2 = (780 x 2) /620

V2 = 2.52 L

Therefore, the volume of the gas when the pressure decreases to 620 mmHg is 2.52 L

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