The work done by a gas molecule in an isolate | vedclass

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(B) The exponent of the exponential function must be dimensionless,so $\frac{x^{2}}{\alpha kT}$ is dimensionless.$[\alpha] = \frac{[x^{2}]}{[kT]} = \f

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$[M L T^{-2}]$

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(B) The exponent of the exponential function must be dimensionless,so $\frac{x^{2}}{\alpha kT}$ is dimensionless.$[\alpha] = \frac{[x^{2}]}{[kT]} = \frac{L^{2}}{M L^{2} T^{-2}} = M^{-1} T^{2}$.Since the exponential term $e^{-\frac{x^{2}}{\alpha kT}}$ is dimensionless,the dimension of work $W$ is given by $[W] = [\alpha][\beta]^{2}$.$[W] = M L^{2} T^{-2}$.Substituting the dimensions: $M L^{2} T^{-2} = (M^{-1} T^{2}) [\beta]^{2}$.$[\beta]^{2} = \frac{M L^{2} T^{-2}}{M^{-1} T^{2}} = M^{2} L^{2} T^{-4}$.Taking the square root: $[\beta] = M L T^{-2}$.

The work done by a gas molecule in an isolated system is given by $W = \alpha \beta^{2} e^{-\frac{x^{2}}{\alpha kT}}$,where $x$ is the displacement,$k$ is the Boltzmann constant,$T$ is the temperature,and $\alpha$ and $\beta$ are constants. Then the dimension of $\beta$ will be:

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