A sample of helium gas occupies 12.9 L at 315 K and 1.20 atm.
This sample contains how many moles of helium?

There are 0,6 moles of helium.

Explanation:

We use the ideal gas formula, we use the gas constant R = 0.082 l atm / K mol and we solve for n (number of moles) of the formula:

PV=nRT -->  n= PV/RT

n= 1,20 atm x 12,9 L/ 0.082 l atm / K mol   x 315 K= 0,599303135 moles

i think so it’s will be b yes i believe its

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