Question

The viscosity of CO2 was measured by comparing its rate of flow through a long narrow tube with that of Ar. For the same pressure differential, the same volume of CO2 passed through the tube in 55 s as that of Ar in 83 s. The viscosity of Ar at 25°C is 208 µP; what is the viscosity of CO2? Estimate the molecular diameter of CO2.

Answer 1

deltaP= 128uLQ/Pid⁴

delaP is pressure drop same and d = diameter of pipe

when pressure drop and d are same.

u= viscosity = 1/Q

u1/u2= Q2/Q1

u1= Viscosity of CO2, u2= 208 Cp.

Q1= V/55 and Q2= V/83

Hence u1/208= 55/83

Hence viscosity = 208*55/83=138CP

Viscosity =138Cp =138*10-2 g/cm.sec=1.38 g/cm.sec= 1.38*10-3 kg/10-2 m/sec= 1.38*10-1 Kg/m.sec= 0.138 kg/m.sec

d ²= (2/3Viscosity)* Sqrt(KTm/Pi³)

Where K= Boltzman constant = 1.38064852 × 10^-23

T= 298.15,

Mass= 44*10-3 kg

When Calculated d= 710 pm