We can start with the definition of coefficient of volume expansion (β), which is given as:
β = (1/V) * (∂V/∂T) * P
Where V is the volume of the substance, T is the temperature, P is the pressure, and ∂V/∂T is the partial derivative of volume with respect to temperature at constant pressure.
Now, let's consider a small temperature change (∆T) and its effect on the density (ρ) of the substance. We can write:
∆ρ = ρ(T + ∆T) - ρ(T)
Using the first order Taylor expansion, we can approximate ρ(T + ∆T) as:
ρ(T + ∆T) ≈ ρ(T) + (∂ρ/∂T)∆T
Where (∂ρ/∂T) is the partial derivative of density with respect to temperature at constant pressure.
Substituting this in the expression for ∆ρ, we get:
∆ρ ≈ ρ(T) + (∂ρ/∂T)∆T - ρ(T)
Simplifying and dividing by ρ(T) and ∆T, we get:
(∆ρ/ρ(T)) ≈ (∂ρ/∂T)/ρ(T) * ∆T
Now, using the ideal gas law, we know that:
PV = nRT
Rearranging, we get:
V = (nRT)/P
Differentiating with respect to temperature at constant pressure, we get:
(∂V/∂T) = (nR/P)
Substituting this in the expression for coefficient of volume expansion (β), we get:
β = (1/V) * (∂V/∂T) * P
= (nR/P) * (1/(nRT/P)) * P
= (1/T)
Therefore, we can write:
(∆ρ/ρ(T)) ≈ -β∆T
Multiplying both sides by ρ(T), we get:
∆ρ ≈ -ρ(T)β∆T
Thus, we have shown that ∆ρ = -ρβ∆T.
Show that ∆ρ = -ρβ∆T
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