Question

A sample of butane gas, C4H10, was slowly heated at constant pressure of .80 bar. The volume of the gas was measured a series of different temperatures and plot of volume vs. temperature was constructed. The slope of the line was .0208 L/K. What was the mass of the sample of butane? Please explain.

Answer 1

We know that PV = nRT

Where P = pressure , V=Volume , n=number of moles , R = gas constant , T = temperature

As P is kept constant & R also constant

V T

V = k T

Which is in the form of y = mx represents a straight line passing through the origin with slope,m = k

Where k = V / T = 0.0208 L/K

Plug the values in PV = nRT   we get

P(V/T) = nR

Where

P = pressure = 80 bar = 80 / 760 atm            Since 1 atm = 760 barr

                                 = 0.105 atm

V/T = 0.0208 L/K

n = number of moles = ?

R = gas constant = 0.0821 Latm/(mol-K)

So (0.105 x 0.0208) = n x 0.0821

                            n = 0.027 moles

Molar mass of butane,C4H10 = (4xAt.mass of C) + (10 x At.mass of H)

                                           = (4x12)+(10x1)

                                           = 58 g/mol

We know that number of moles , n = mass/molar mass

So mass of butane, m = Molar mass x number of moles

                                  = 58 (g/mol) x 0.027 mol

                                  = 1.547 g

Therefore the mass of butane is 1.547 g