The specific volume of water is given quite accurately by the empirical relation
ln(v) = −6.7081 + 1.01257 ln(T) +
280.663/T
Use this relation to derive an equation for the coefficient of thermal expansion � and determine
the temperature at which this quantity is equal to zero. What is the significance of
this temperature?
To derive an equation for the coefficient of thermal expansion (α), we can use the definition of α:
α = (1/v) * (∂v/∂T)
First, let's differentiate the given equation ln(v) = −6.7081 + 1.01257 ln(T) + 280.663/T with respect to T:
d/dT [ln(v)] = d/dT [-6.7081 + 1.01257 ln(T) + 280.663/T]
Using the chain rule, we can differentiate each term separately:
(1/v) * dv/dT = 1.01257 * (1/T) + (-280.663/T^2)
Now, we can rewrite the equation for α by replacing dv/dT in terms of v:
α = (1/v) * (1/v) * (1.01257 * (1/T) + (-280.663/T^2))
Simplifying further:
α = 1/v^2 * (1.01257 * (1/T) + (-280.663/T^2))
To determine the temperature at which α is equal to zero, we set α = 0 and solve for T:
0 = 1/v^2 * (1.01257 * (1/T) + (-280.663/T^2))
Since v cannot be zero, we can ignore the first term on the right side of the equation. We are left with:
0 = 1.01257 * (1/T) + (-280.663/T^2)
Multiply through by T^2 to eliminate the denominators:
0 = 1.01257 * T - 280.663
Solving for T:
T = 280.663 / 1.01257
T ≈ 277.4 K
The significance of this temperature is that it represents the point at which the coefficient of thermal expansion for water is zero. This temperature is known as the temperature of maximum density for water. At this temperature, water reaches its highest density, which has various implications in fields like climatology and oceanography. Water expands when it is heated or cooled from this temperature.