Question
Calculate the temperature at which the root mean square velocity of nitrogen molecules will be equal to $8km-s^{-1}$.
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Answer
Given, r.m.s. velocity, $C = 8km-s^{-1} = 8 \times 10^5cm-s^{-1}$ Molar gas constant $= R = 8.31 \times 10^7erg ~mol^{-1}K^{-1}$ Molecular weight of nitrogeru M = 28 Let T be the required temperature Then, $\text{C}=\sqrt{\frac{3\text{RT}}{\text{M}}}$ or $\text{C}^2=\frac{3\text{RT}}{\text{M}}$ or $\text{T}=\frac{\text{MC}^2}{3\text{R}}$$=\frac{28\times(8\times10^5)^2}{3\times8.31\times10^7}\text{K}=71881\text{K}$
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