
Show that volume expansion coefficient is three times the linear expansion coefficient of a solid.
Answer
560.1k+ views
Hint:As we know that thermal expansion is the increase or decrease in the length, area or volume of a body due to change in temperature. When any body or object is heated up then the change happens in the expansion coefficients.
Formula used:
Change in volume due to thermal expansion is-
\[\Delta V = \gamma V\Delta T\]
Here \[\gamma \] is the coefficient of volume expansion and \[\gamma \] \[ \approx 3\alpha \]. Here \[\alpha \] is the linear expansion coefficient.
Complete step by step answer:
Linear thermal expansion is defined as the fractional increase in length whereas the volume expansion is defined as the fractional increase in volume per unit rise in temperature. Now, consider one cuboid whose initial dimensions are \[{l_1},{l_2},{l_3}\] and initial volume is
\[V = {l_1}{l_2}{l_3}\]
Final dimensions of cuboid are-
\[{l_1}' = {l_1}\left( {1 + \alpha \Delta T} \right)\]
\[{l_2}' = {l_2}\left( {1 + \alpha \Delta T} \right)\]
\[{l_3}' = {l_3}\left( {1 + \alpha \Delta T} \right)\]
So, final volume will be-
\[V' = {l_1}'{l_2}'{l_3}'\]
\[V' = {l_1}\left( {1 + \alpha \Delta T} \right){l_2}\left( {1 + \alpha \Delta T} \right){l_3}\left( {1 + \alpha \Delta T} \right)\]
So,\[V' = {l_1}{l_2}{l_3}{\left( {1 + \alpha \Delta T} \right)^3}\]
Here, the linear expansion coefficient of linear expansion \[\alpha \] is very small. So, by using binomial concept, \[{\left( {1 + \alpha \Delta T} \right)^3} \approx 1 + 3\alpha \Delta T\]
\[V' = V(1 + 3\alpha \Delta T) = V(1 + \gamma \Delta T)\]
So, \[\gamma \]\[ = 3\alpha \]
Hence, coefficient of volume expansion coefficient is three times of the linear expansion coefficient.
Note:Remember to take care of the approximations that are mentioned in the questions. Also there are three types of thermal expansion- Linear expansion is the change in length due to heating, Volume expansion is the change in volume due to heating and Area expansion is the change in area due to heating. So, different formulas will be used for all of them.
Formula used:
Change in volume due to thermal expansion is-
\[\Delta V = \gamma V\Delta T\]
Here \[\gamma \] is the coefficient of volume expansion and \[\gamma \] \[ \approx 3\alpha \]. Here \[\alpha \] is the linear expansion coefficient.
Complete step by step answer:
Linear thermal expansion is defined as the fractional increase in length whereas the volume expansion is defined as the fractional increase in volume per unit rise in temperature. Now, consider one cuboid whose initial dimensions are \[{l_1},{l_2},{l_3}\] and initial volume is
\[V = {l_1}{l_2}{l_3}\]
Final dimensions of cuboid are-
\[{l_1}' = {l_1}\left( {1 + \alpha \Delta T} \right)\]
\[{l_2}' = {l_2}\left( {1 + \alpha \Delta T} \right)\]
\[{l_3}' = {l_3}\left( {1 + \alpha \Delta T} \right)\]
So, final volume will be-
\[V' = {l_1}'{l_2}'{l_3}'\]
\[V' = {l_1}\left( {1 + \alpha \Delta T} \right){l_2}\left( {1 + \alpha \Delta T} \right){l_3}\left( {1 + \alpha \Delta T} \right)\]
So,\[V' = {l_1}{l_2}{l_3}{\left( {1 + \alpha \Delta T} \right)^3}\]
Here, the linear expansion coefficient of linear expansion \[\alpha \] is very small. So, by using binomial concept, \[{\left( {1 + \alpha \Delta T} \right)^3} \approx 1 + 3\alpha \Delta T\]
\[V' = V(1 + 3\alpha \Delta T) = V(1 + \gamma \Delta T)\]
So, \[\gamma \]\[ = 3\alpha \]
Hence, coefficient of volume expansion coefficient is three times of the linear expansion coefficient.
Note:Remember to take care of the approximations that are mentioned in the questions. Also there are three types of thermal expansion- Linear expansion is the change in length due to heating, Volume expansion is the change in volume due to heating and Area expansion is the change in area due to heating. So, different formulas will be used for all of them.
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