Express volume expansivity in terms of density

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SUMMARY

The discussion focuses on expressing volume expansivity (B) in terms of density (ρ) and its partial derivatives. The key equation derived is B = (-1/ρ)(dρ/dT), which results from substituting volume (V = m/ρ) into the expansivity equation B = (1/V)(dV/dT). The manipulation of the differential equation involves implicit differentiation, leading to the conclusion that mass (m) remains constant with temperature changes, simplifying the relationship between volume expansivity and density.

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  • Understanding of volume expansivity (B) and its definition
  • Familiarity with the concept of density (ρ) and its relationship to mass (m) and volume (V)
  • Knowledge of partial derivatives and implicit differentiation techniques
  • Basic principles of thermodynamics related to temperature effects on density
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matthewsyq1995
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Homework Statement


Express volume expansivity (B) in terms of density (ρ) and its partial derivatives

Homework Equations


B = (1/V) (dV/dT)
V = m/ρ

The Attempt at a Solution


I have only managed to substitute m/ρ into the expansivity equation.
Don't really understand how to manipulate the differential equation into (-1/ρ)(dρ/dT)

Especially the part where d[(m/ρ)]/(m/ρ) equals (-1/ρ)dρ
 
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What are the properties of "m?"
 
Given the equations you provided, we substitute V=m/ρ to find that β= (ρ/m)*(d/dT(m/ρ)). Then, take the derivative of 1/ρ with respect to temperature, so we don't just end up with -m/(ρ^2), we end up with -m/(ρ^2)*(dρ/dT). Mass is of course constant, and does not (noticeably) change with temperature. (Think back to implicit differentiation where we differentiate y^2 with respect to x to get 2y*y'). Now, we have β= (ρ/m)*(-m/[ρ][/2]*(dρ/dT)) which, when we cancel out a ρ and the m term, gives us β=(-1/ρ)(dρ/dT).
 
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