In a typical combustion engine,the work done | vedclass

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(B) The term $kT$ represents energy,so its dimension is $[ML^{2}T^{-2}]$.The exponent $\frac{-\beta x^{2}}{kT}$ must be dimensionless.Therefore,$[\bet

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$[M^{0}LT^{0}]$

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(B) The term $kT$ represents energy,so its dimension is $[ML^{2}T^{-2}]$.The exponent $\frac{-\beta x^{2}}{kT}$ must be dimensionless.Therefore,$[\beta][x^{2}] = [kT] \implies [\beta][L^{2}] = [ML^{2}T^{-2}]$.This gives $[\beta] = [MT^{-2}]$.The dimension of work $W$ is $[ML^{2}T^{-2}]$.Given $W = \alpha^{2} \beta e^{\frac{-\beta x^{2}}{kT}}$,and since the exponential term is dimensionless,we have $[W] = [\alpha^{2}][\beta]$.$[ML^{2}T^{-2}] = [\alpha^{2}][MT^{-2}]$.$[\alpha^{2}] = \frac{[ML^{2}T^{-2}]}{[MT^{-2}]} = [L^{2}]$.Therefore,$[\alpha] = [L] = [M^{0}LT^{0}]$.

In a typical combustion engine,the work done by a gas molecule is given by $W = \alpha^{2} \beta e^{\frac{-\beta x^{2}}{kT}}$,where $x$ is the displacement,$k$ is the Boltzmann constant,and $T$ is the temperature. If $\alpha$ and $\beta$ are constants,the dimensions of $\alpha$ will be:

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