To solve for the apparent expansivity of mercury in a glass container, we need to understand the relationships between the properties given.
The apparent expansivity [tex]\(\alpha_{app}\)[/tex] is determined by the difference between the cubic expansivity of the mercury and three times the linear expansivity of the glass container that holds it. This is because the glass container will expand linearly in three dimensions, affecting the volume measurement of the mercury.
Given:
- Cubic expansivity of mercury, [tex]\(\alpha_m = 1.8 \times 10^{-4} \, K^{-1}\)[/tex]
- Linear expansivity of glass, [tex]\(\alpha_g = 8.0 \times 10^{-6} \, K^{-1}\)[/tex]
The formula for apparent expansivity is given by:
[tex]\[ \alpha_{app} = \alpha_m - 3 \alpha_g \][/tex]
Now substitute in the given values:
[tex]\[ \alpha_{app} = 1.8 \times 10^{-4} - 3 \times 8.0 \times 10^{-6} \][/tex]
Calculate [tex]\(3 \times 8.0 \times 10^{-6}\)[/tex]:
[tex]\[ 3 \times 8.0 \times 10^{-6} = 24.0 \times 10^{-6} = 2.4 \times 10^{-5} \][/tex]
Now subtract [tex]\(2.4 \times 10^{-5}\)[/tex] from [tex]\(1.8 \times 10^{-4}\)[/tex]:
[tex]\[ \alpha_{app} = 1.8 \times 10^{-4} - 2.4 \times 10^{-5} \][/tex]
Convert both terms to a common exponent (for easy subtraction):
[tex]\[ 1.8 \times 10^{-4} = 18.0 \times 10^{-5} \][/tex]
Now subtract:
[tex]\[ \alpha_{app} = 18.0 \times 10^{-5} - 2.4 \times 10^{-5} = 15.6 \times 10^{-5} \][/tex]
Convert back to the standard form with the negative exponent:
[tex]\[ \alpha_{app} = 1.56 \times 10^{-4} \, K^{-1} \][/tex]
Thus, the apparent expansivity of mercury in a glass container is:
[tex]\[ \alpha_{app} = 1.56 \times 10^{-4} \, K^{-1} \][/tex]
