In a typical combustion engine the work done by a gas molecule is given by W = <math><mrow><msup><mrow><mi>α</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>β</mi><msup><mrow><mi>e</mi></mrow><mrow><mo>−</mo><mfrac><mrow><mi>β</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>k</mi><mi>T</mi></mrow></mfrac></mrow></msup><mo>,</mo></mrow></math> where x is the displacement, k is the Boltzmann constant and T is the temperature. If and are constants, dimensions of will be

W=α2β e−−βx2kT [βx2]= [kT]= [ML2T−2]β=MT−2 [W]= [α2β]⇒ [ML2T−2]= [α2] [MT−2]⇒ [α2]= [L2]⇒ [α]= [L]

In a typical combustion engine the work done by a gas molecule is given by W = <math><mrow><msup><mrow><mi>α</mi></mrow><mrow><mn>2</mn></mrow></msup><mi>β</mi><msup><mrow><mi>e</mi></mrow><mrow><mo>−</mo><mfrac><mrow><mi>β</mi><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>k</mi><mi>T</mi></mrow></mfrac></mrow></msup><mo>,</mo></mrow></math>&nbsp;where x is the displacement, k is the Boltzmann constant and T is the temperature. If and are constants, dimensions of will be

Ask & Answer Question

Option 1 -

$[M L T^{- 1}]$

Option 2 -

$[M^{0} L T^{0}]$

Option 3 -

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

Option 4 -

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

2 Views | Posted a month ago

Answered by

Payal Gupta | Contributor-Level 10

a month ago

Correct Option - 2

Detailed Solution:

$W = α^{2} β e^{- \frac{- β x^{2}}{k T}}$

$[β x^{2}] = [k T] = [M L^{2} T^{- 2}]$

$β = M T^{- 2}$

$[W] = [α^{2} β] \Rightarrow [M L^{2} T^{- 2}] = [α^{2}] [M T^{- 2}] \Rightarrow [α^{2}] = [L^{2}] \Rightarrow [α] = [L]$

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