Heat And Thermodynamics Ques 1
1 The number density of molecules of a gas depends on their distance $r$ from the origin as, $n(r)=n_0 e^{-a r^4}$. Then, the total number of molecules is proportional to
(2019 Main, 12 April II)
(a) $n_0 \alpha^{-3 / 4}$
(b) $\sqrt{n_0} \alpha^{1 / 2}$
(c) $n_0 \alpha^{1 / 4}$
(d) $n_0 \alpha^{-3}$
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Answer:
Correct Answer: 1.( a )
Solution:
Formula:
Total Translational K.E. Of Gas
$\begin{aligned} & \text { 1. Number density of gas molecules, } \\ & n=\frac{\text { Number of molecules }}{\text { Volume of gas }} \\ & \Rightarrow \text { Number of molecules, } \\ & N=n \times \text { volume of gas }\end{aligned}$
Volume of shell of differentiable thickness $(d r), d V=$ surface area $\times$ thickness $=4 \pi r^2 d r$
Now, number of molecules in this shell is
$ d N=n(r) \cdot d V=n_0 e^{-\omega \omega^4} \cdot 4 \pi r^2 d r $
So, total number of molecules present in given volume (extending from $r=0$ to $r=\infty$ ) is $ N=\int_0^{\infty} n(r) \cdot d V=\int_0^{\infty} n_0 e^{-a r^4} \cdot 4 \pi r^2 d r $
$ =\int_0^{\infty} 4 \pi n_0 e^{-\alpha r^4} \cdot r^2 d r $
Here, we take
$ \begin{aligned} & \alpha r^4=t \Rightarrow r=t^{\frac{1}{4}} \cdot \alpha^{\frac{-1}{4}} \\ & \Rightarrow \quad 4 \alpha r^3 d r=d t \\ & \Rightarrow \quad r^2 d r=\frac{d t}{4 \alpha}=\frac{d t}{4 \alpha t^{\frac{1}{4}} \cdot \alpha^{-\frac{1}{4}}}=\frac{d t}{4 \alpha^{3 / 4} t^{\frac{1}{4}}} \\ & \end{aligned} $
Also, when $r=0, t=0$ and when $r=\infty, t=\infty$ substituting in Eq. (i), we get
$ \begin{aligned} & N=\int_0^{\infty} 4 \pi n_0 e^{-t} \cdot \frac{d t}{4 \alpha^{\frac{3}{4}} t^{\frac{1}{4}}} \\ & N=\pi \alpha^{-\frac{3}{4}} \cdot n_0 \cdot \int_0^{\infty} e^{-t} \cdot t^{-\frac{1}{4}} d t \end{aligned} $
As value of definite integral $\int_0^{\infty} e^{-t} \cdot t^{-\frac{1}{4}} d t$ is a constant $(=k$ let), we have
$ N=\pi k n_0 \alpha^{-\frac{3}{4}} \Rightarrow N \propto n_0 \alpha^{-\frac{3}{4}} $
BETA
