Question #61171

Calculate the coefficient of viscosity of hydrogen at STP. Take
r = 8.90×10−2 kg m−3, l = 2×10−7m and =1.38×10−23 JK−1.

Expert's answer

Answer on Question #61171-Physics-Molecular Physics | Thermodynamics

Calculate the coefficient of viscosity of hydrogen at STP. Take

r=8.90×102kgm3,l=2×107mr = 8.90 \times 10^{-2} \, \text{kg} \, \text{m}^{-3}, l = 2 \times 10^{-7} \, \text{m} and =1.38×1023JK1= 1.38 \times 10^{-23} \, \text{JK}^{-1} .

Solution


η=13ρvλ\eta = \frac {1}{3} \rho v \lambdav=3kTmv = \sqrt {\frac {3 k T}{m}}


At t=273kt = 273k and volume 22.4103m322.4 \cdot 10^{-3} \, m^3 .


m=ρVNa=8.9010222.41036.021023=3.311027kgm = \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 gv=31.3810232733.311027=1848ms.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}.η=138.9010218482107=1.09105kgms\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}


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