If you are given an ideal gas with pressure (p) = 259,392.00 pa and temperature (t) = 2.00 oc of 1 mole argon gas in a volume of 8.8 dm3 , calculate r to the correct number of significant figures and units under given condition

R = 0.064 dm³ atm K⁻¹ mol⁻¹

Explanation:

Data Given:

volume of gas V = 8.8 dm³

no. of mole of gas (n) = 1 mole

Pressure P = 259,392.00 Pa

Convert Pascal (Pa) to atm (atmospheric pressure)

As,

101,325 Pa = 1 atm

So,

259,392.00 Pa = 2 atm

Then Pressure (P) = 2 atm

Temperature T = 2.00 °C

change the temperature from °C to K

As to convert °C to K the below formula used

                       0°C + 273.15 = 273.15K

So, for 2 °C

                      2°C + 273.15 = 275.15 K

So,

Temperature T = 275.15 K

ideal gas constant = ?

formula used for Ideal gases

                           PV = nRT

as we have to find R of the gas:

we will rearrange the ideal gas equation as below:

                           R = PV / nT (1)

Put value in equation (1)

                             R = 2atm x 8.8 dm³ / 1 mole x 275.15 K

                             R = 17.6 atm. dm³ / 275.15 mol. K

                             R = 0.064 dm³ atm K⁻¹ mol⁻¹

So the value of R is 0.064 dm³ atm K⁻¹ mol⁻¹

and the unit of R (ideal gas constant) is dm³ atm K⁻¹ mol⁻¹

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