Answer on Question #61171-Physics-Molecular Physics | Thermodynamics
Calculate the coefficient of viscosity of hydrogen at STP. Take
r = 8.90 × 1 0 − 2 kg m − 3 , l = 2 × 1 0 − 7 m r = 8.90 \times 10^{-2} \, \text{kg} \, \text{m}^{-3}, l = 2 \times 10^{-7} \, \text{m} r = 8.90 × 1 0 − 2 kg m − 3 , l = 2 × 1 0 − 7 m and = 1.38 × 1 0 − 23 JK − 1 = 1.38 \times 10^{-23} \, \text{JK}^{-1} = 1.38 × 1 0 − 23 JK − 1 .
Solution
η = 1 3 ρ v λ \eta = \frac {1}{3} \rho v \lambda η = 3 1 ρ v λ v = 3 k T m v = \sqrt {\frac {3 k T}{m}} v = m 3 k T
At t = 273 k t = 273k t = 273 k and volume 22.4 ⋅ 1 0 − 3 m 3 22.4 \cdot 10^{-3} \, m^3 22.4 ⋅ 1 0 − 3 m 3 .
m = ρ V N a = 8.90 ⋅ 1 0 − 2 ⋅ 22.4 ⋅ 1 0 − 3 6.02 ⋅ 1 0 23 = 3.31 ⋅ 1 0 27 k g m = \frac {\rho V}{N _ {a}} = \frac {8 . 9 0 \cdot 1 0 ^ {- 2} \cdot 2 2 . 4 \cdot 1 0 ^ {- 3}}{6 . 0 2 \cdot 1 0 ^ {2 3}} = 3. 3 1 \cdot 1 0 ^ {2 7} k g m = N a ρ V = 6.02 ⋅ 1 0 23 8.90 ⋅ 1 0 − 2 ⋅ 22.4 ⋅ 1 0 − 3 = 3.31 ⋅ 1 0 27 k g v = 3 ⋅ 1.38 ⋅ 1 0 − 23 ⋅ 273 3.31 ⋅ 1 0 27 = 1848 m s . v = \sqrt {\frac {3 \cdot 1 . 3 8 \cdot 1 0 ^ {- 2 3} \cdot 2 7 3}{3 . 3 1 \cdot 1 0 ^ {2 7}}} = 1 8 4 8 \frac {m}{s}. v = 3.31 ⋅ 1 0 27 3 ⋅ 1.38 ⋅ 1 0 − 23 ⋅ 273 = 1848 s m . η = 1 3 ⋅ 8.90 ⋅ 1 0 − 2 ⋅ 1848 ⋅ 2 ⋅ 1 0 − 7 = 1.09 ⋅ 1 0 − 5 k g m s \eta = \frac {1}{3} \cdot 8. 9 0 \cdot 1 0 ^ {- 2} \cdot 1 8 4 8 \cdot 2 \cdot 1 0 ^ {- 7} = 1. 0 9 \cdot 1 0 ^ {- 5} \frac {k g}{m s} η = 3 1 ⋅ 8.90 ⋅ 1 0 − 2 ⋅ 1848 ⋅ 2 ⋅ 1 0 − 7 = 1.09 ⋅ 1 0 − 5 m s k g
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