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  • If the number density of molecules of a gas depends upon their distance x fromn one origin.' <img alt="n(x)=n \cdot e^{-\alpha x^3}" src="https://entrancecorner.oncodecogs.com/gif.lat

If the number density of molecules of a gas depends upon their distance x fromn one origin.'

n(x)=n \cdot e^{-\alpha x^3}. Then find the total number of molecules is directly proportional to-

Option: 1

\mathrm{\frac{4 \pi n_0}{3 x}\left(1-e^{-2 x^3}\right)}


Option: 2

\mathrm{4 \pi n \cdot \alpha}


Option: 3

\mathrm{\left(e^{-\alpha x^3}+1\right)}


Option: 4

none of these.


Answers (1)

best_answer

\begin{aligned} N & =\int \rho d v=\int_0^x n_0 e^{-\alpha x^3} 4 \pi x^2 d x \\ & =n_0 4 \pi \int_0^x x^2 e^{-2 x^3} d x . \end{aligned}

Put \: x^3=t\\\\\ \begin{aligned} 3 x^2 d x & =d t \\ d x & =\frac{d t}{3 x^2} \end{aligned}

\begin{aligned} N & =\operatorname{noun}_0^x x e^{-\alpha t} \frac{d t}{3 x y} \\ & =\frac{n_0 \frac{4 \pi}{3}}{\int_0^x} \int_0^{-\alpha t} \\ & =\frac{417 n_0}{3}\left[\frac{e^{-\alpha t}}{-\alpha}\right]_0^x \end{aligned}

\begin{aligned} & =\frac{413 n_0}{3}\left[\frac{e^{-\alpha x^3}}{=\alpha}\right]_0^x \\ & =\frac{4 \pi n_0}{3}\left[\frac{e^{-x^3}}{-\alpha}-\frac{1}{-\alpha}\right] \\ & =\frac{4 \pi n_0}{3}\left[-\frac{e^{-\alpha x^3}}{2}+\frac{1}{\alpha}\right] \end{aligned}

=\frac{4 \pi n_0}{3} \alpha\left[1-e^{-\alpha x^3}\right]

 

 

 

Posted by

seema garhwal

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