Question

A quantity of CO gas occupies a volume of 0.58 Lat 1.2 atm and 288 K . The pressure of the gas is lowered and its temperature is raised until its volume is 3.6 L .

Find the density of the CO under the new conditions.

Answer 1

Answer – We are given, P1 = 1.2 atm , V1 0.58 L , T = 288 K

V2 = 3.6 L , P2 = ?

We know,

P1V1 = P2V2

So, P2 = 1.2 atm * 0.58 L / 3.6 L

            = 0.193 atm

We also know

V1/T1 = V2/T2

So, T2 = V2 * T1 / V1

            = 3.6 L * 288 K / 0.58 L

            = 1787.6 K

So new pressure P = 0.193 atm and temp. T = 1787.6 K

So, the density of CO

Density of CO = PM/RT

                      = 0.193 atm * 28.01 g.mol^-1 / 0.0821 L.atm.mol^-1.K^-1*1787.6 K

                      = 0.0369 g/L