At this pressure, how many molecules of air would there be in a 20 ∘c experimental chamber with a volume of 0.020 m3 ?

Explanation: The lowest pressure in a laboratory is 4.0×10^-11Pa

Using Ideal gas equation

PV = nRT

P= 4.0×10^-11Pa

V= 0.020m^3

T= 20+273= 293k

n=number of moles = m/A

Where m is the number of molecules and A is the Avogradro's number=6.02×10²³/mol

R=8.314J/(mol × K)

PV= m/A(RT)

4.0×10^-11 ×0.020 = m/6.02×10²³(8.314×293)

m = 4.0×10^-11×0.020×6.02×10^23 / (8.314×293)

m = 1.98×10^8 molecules

Therefore,the number of molecules is 1.98×10^8

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a gas, having a molecular weight of 13 kg/kmol and a kinematic viscosity of 0.25 cm2/s, flows through a pipe 0.25 m internal diameter and 5 km long at the rate of 0.4 m3/s and is delivered at atmospheric pressure.