Question

H2 and N2 are maintained under identical conditions of temperature and pressure. Find the ratio of coefficients of viscosity. Diameter of H2 and N2 molecules are 2.5*10^(-8) cm and 3.5*10^(-8) cm respectively.​

Answer 1

Answer:

Explanation:

The diameter will be 3.5×1p(-8)

Answer 2

Given,

Diameter of H₂ = 2.5 x $10^{-8}$ cm

hence, radius of H₂ = 1.25 x $10^{-8}$ cm ≈ 1.3 x $10^{-8}$ cm

Diameter of N$_{2}$ = 3.5 x $10^{-8}$cm

hence, radius of N$_{2}$ = 1.75 x $10^{-8}$cm≈ 1.8 x $10^{-8}$ cm

To Find,

the ratio of coefficients of viscosity.

Solution,

Viscosity = $\frac{F}{A} =$ η$\frac{du}{dt}$

( v = velocity gradient is independent of radius)

now,

η∝ $\frac{1}{A}$∝$\frac{1}{r^{2} }$

therefore,

η₁ = $(\frac{1.8 X 1.8}{1.3 X 1.3} )^{2}$ = 1.91

η₂

∴Hence,  the ratio of coefficients of viscosity is 1.91.