Question

Calculate the fraction of $N_2$ molecules at 101.325 kpa and 300 K whose speeds are in the range of
$u_{mp}-0.005u_{mp}$ to $u_{mp}+0.005u_{mp}$:

• A. $4.15\times {10}^{-3}$
• B. $8.303\times {10}^{-3}$
• C. $16.606\times {10}^{-4}$
• D. None of these

Answer 1

Answer: (B)

$8.303\times {10}^{-3}$

$u_{sp}=\sqrt{\frac{2RT}{M}}={\left\{\frac{2\left(8.314J{K}^{-1}mo{l}^{-1}\right)\left(300K\right)}{(0.028kgmo{l}^{-1}}\right\}}^{1/2}$

$=422.09J^{1/2}k{g}^{-1/2}=422.09m{s}^{-1}$

Since $0.005u_{mp}=\left(0.005\right)(422.09m{s}^{-1}$

therefore du $=(u_{mp}+0.005u_{mp})-(u_{mp}-0.005u_{mp})=4.22$

Maxxwell distribution law states

$\frac{dN}{N}=4\pi {\left(\frac{M}{2\pi RT}\right)}^{3/2}{u}^{2}exp\left(M{u}^2/2RT\right)du$

${\left(\frac{M}{2\pi RT}\right)}^{3/2}={\left\{\frac{\left(0.028kgmo{l}^{-1}\right)}{2\times 3.14\times \left(8.314J{K}^{-1}mo{l}^{-1}\right)\left(300k\right)}\right\}}^{3/2}$

$exp(-\frac{M{u}^2}{2RT})=exp\left\{\frac{\left(0.028kgmo{l}^{-1}\right)(422.09m{s}^{-1}{)}^2}{2\left(8.314J{K}^{-1}mo{l}^{-1}\right(300K)}\right\}$

$\frac{dN}{N}=4\times 3.14\times (2.390\times {10}^{-9}{m}^{-3}{s}^2)\left(422.09m{s}^{-1}{)}^2\right(0.3679\left)\right(4.22m{s}^{-1})$

$=8.303\times {10}^{-3}$