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Density is mass divided by volume. If the volume \(V\) is temperature dependent, so is the density \(\rho\). Show that the change in density \(\Delta \rho\) with change in temperature \(\Delta T\) is given by $$\Delta \rho=-\beta \rho \Delta T$$ where \(\beta\) is the coefficient of volume expansion. Explain the minus sign.

Short Answer

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The change in density with temperature \(\Delta \rho\) is given by \(\Delta \rho=-\beta \rho \Delta T\). The minus sign indicates that the density decreases as the temperature increases, assuming a positive coefficient of volume expansion.

Step by step solution

01

Understanding the Basics

The basic formula for density is \(\rho=\frac{m}{V}\) where \(m\) is mass, \(V\) is volume, and \(\rho\) is density. Since the mass \(m\) does not change with temperature, we only need to consider volume \(V\) which is affected by temperature.
02

Incorporating Coefficient of Volume Expansion

The formula that incorporates the change in volume with temperature is \(V=V_0(1+\beta \Delta T)\) where \(V_0\) is the initial volume, \(\beta\) is the coefficient of volume expansion, and \(\Delta T\) is the change in temperature. If we substitute this back into the original density equation, we get \(\rho=\frac{m}{V_0(1+\beta \Delta T)}\). In other words, density decreases when volume increases due to increase in temperature.
03

Finding the Change in Density with Respect to Temperature

To find the change in density \(\Delta \rho\) with change in temperature \(\Delta T\), we differentiate the density equation with respect to \(T\). This gives us \(\Delta \rho=-\frac{m \beta}{V_0(1+\beta \Delta T)^2}\Delta T=-\beta \rho \Delta T\).
04

Explaining the Minus Sign

The negative sign in \(\Delta \rho=-\beta \rho \Delta T\) signifies that the density decreases as the temperature increases, given a positive value of \(\beta\). This is because as temperature increases, volume increases leading to decrease in density.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coefficient of Volume Expansion
The coefficient of volume expansion, denoted as \(\beta\), is a term that describes how the volume of a material changes with temperature. It is a measure that tells us how much a given unit of volume will expand or contract when the temperature changes by one degree. This concept is crucial for understanding the behavior of substances under temperature fluctuations.
  • It is expressed in units of reciprocal temperature (e.g., 1/°C or 1/K).
  • The mathematical relationship can be seen in the formula \(V = V_0(1 + \beta \Delta T)\), showing how volume expands as temperature increases.
  • A positive \(\beta\) means the substance expands when heated and contracts when cooled.
Understanding this coefficient helps us anticipate and calculate changes in materials' properties, such as density and structural integrity, in response to temperature changes.
Temperature Dependence
Temperature has a profound effect on the physical properties of materials, including their volume and density. As temperature changes, the kinetic energy of the molecules in a substance changes, causing them to vibrate more vigorously and occupy more space.
This temperature dependence can lead to:
  • Increased volume as temperature rises, leading to decreased density.
  • Decreased volume as temperature falls, leading to increased density.
The formula \(V = V_0(1 + \beta \Delta T)\) captures how volume depends on temperature and utilizes the coefficient of volume expansion \(\beta\). This formula outlines that any change in temperature \(\Delta T\) results in a proportional change in volume, and subsequently, affects the density.
Change in Density
Density is the measure of mass per unit volume. It follows that if the volume of a substance changes while the mass remains constant, the density must also change. The mathematical expression for density is \(\rho = \frac{m}{V}\).
When there is a change in temperature:\
  • The coefficient of volume expansion \(\beta\) influences the volume change, which affects the density.
  • The change in density is given by \(\Delta \rho = -\beta \rho \Delta T\).
  • The negative sign indicates an inverse relationship: as temperature increases (positive \(\Delta T\)), the density decreases, assuming \(\beta\) is positive.
This relationship highlights that density decreases with temperature rise because the volume expands while the mass remains constant. Such understanding is crucial in fields like material science, engineering, and physical sciences where temperature-induced density changes can significantly impact material properties and system behaviors.

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Most popular questions from this chapter

Oxygen gas having a volume of \(1130 \mathrm{~cm}^{3}\) at \(42.0^{\circ} \mathrm{C}\) and a pressure of \(101 \mathrm{kPa}\) expands until its volume is \(1530 \mathrm{~cm}^{3}\) and its pressure is \(106 \mathrm{kPa}\). Find \((a)\) the number of moles of oxygen in the system and ( \(b\) ) its final temperature.

A quantity of ideal gas at \(12.0^{\circ} \mathrm{C}\) and a pressure of \(108 \mathrm{kPa}\) occupies a volume of \(2.47 \mathrm{~m}^{3}\). (a) How many moles of the gas are present? \((b)\) If the pressure is now raised to \(316 \mathrm{kPa}\) and the temperature is raised to \(31.0^{\circ} \mathrm{C}\), how much volume will the gas now occupy? Assume there are no leaks.

An aluminum cup of \(110 \mathrm{~cm}^{3}\) capacity is filled with glycerin at \(22^{\circ} \mathrm{C}\). How much glycerin, if any, will spill out of the cup if the temperature of the cup and glycerin is raised to \(28^{\circ} \mathrm{C}\) ? (The coefficient of volume expansion of glycerin is \(5.1 \times\) \(\left.10^{-4} / \mathrm{C}^{\circ} .\right)\)

An air bubble of \(19.4 \mathrm{~cm}^{3}\) volume is at the bottom of a lake \(41.5 \mathrm{~m}\) deep where the temperature is \(3.80^{\circ} \mathrm{C}\). The bubble rises to the surface, which is at a temperature of \(22.6^{\circ} \mathrm{C}\). Take the temperature of the bubble to be the same as that of the surrounding water and find its volume just before it reaches the surface.

Repeat Exercise 1, except choose the new temperature scale \(\mathrm{Q}\) so that absolute zero is \(0^{\circ} \mathrm{Q}\) and \(T_{\mathrm{bp} \text { , water }}-T_{\text {mp, water }}=\) \(100 \mathrm{Q}^{\circ} .(a)\) What is the conversion formula from Celsius to Q? (b) What is \(T_{\text {bp, water }}\) and \(T_{\text {mp,water }}\) in Q? ( \(c\) ) This scale actually exists. What is the official name?

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