Problem 2 - Chemical Principles in the Laboratory

Question

In this experiment a student found that when she increased the temperature of a 544 mL sample of air from \(22.8^{\circ} \mathrm{C}\) to \(33.6^{\circ} \mathrm{C},\) the pressure of the air went from \(1012 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\) up to \(1049 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\). Since the air expands linearly with temperature, the equation relating \(P\) to \(t\) is of the form: $$\boldsymbol{P}=m t+\boldsymbol{b}$$ where \(m\) is the slope of the line and \(b\) is a constant. a. What is the slope of the line? (Find the change in \(P\) divided by the change in \(t\).) \(m=\)_______\(\mathrm{cm} \mathrm{H}_{2} \mathrm{O} /^{\circ} \mathrm{C}\) b. Find the value of \(b\). (Substitute known values of \(P\) and \(t\) into Equation 9 and solve for \(b\).) \(b=\)______ \(\mathbf{c m} \mathbf{H}_{2} \mathbf{O}\) c. Express Equation 9 in terms of the values of \(m\) and \(b.\) $$ P= $$ d. At what temperature \(t\) will \(P\) become zero? \(P=0\) at \(t=\) ______ \(^{\circ} \mathrm{C}=t_{\circ}=-k\) (Please note that you are unlikely to get results anywhere near this good when actually carrying out this experiment, as it involves a very large extrapolation.) e. The temperature in Part (d) is the absolute zero of temperature. Lord Kelvin suggested that we set up a scale on which that temperature is \(0 \mathrm{K}\). On that scale, \(T=t+k .\) Show that, on the Kelvin scale, your equation reduces to \(P=m T\)

Step-by-Step Solution

Answer

a) \( m = 3.43 \, \text{cm H}_2\text{O}/^{\circ} \text{C} \), b) \( b = 933.796 \, \text{cm H}_2\text{O} \), c) \( P = 3.43t + 933.796 \), d) \( t \approx -272.31^{\circ} \text{C} \), e) On Kelvin scale \( P = mT \).

Step 1: Calculate the Change in Pressure

To find the slope of the line, we first need to calculate the change in pressure (\( P \)) which is the difference between the final pressure and the initial pressure. Thus:\[\Delta P = P_{final} - P_{initial} = 1049 \, \text{cm H}_2\text{O} - 1012 \, \text{cm H}_2\text{O} = 37 \, \text{cm H}_2\text{O}\]

Step 2: Calculate the Change in Temperature

Next, we calculate the change in temperature (\( t \)) which is the difference between the final and initial temperatures:\[\Delta t = t_{final} - t_{initial} = 33.6^{\circ} \text{C} - 22.8^{\circ} \text{C} = 10.8^{\circ} \text{C}\]

Step 3: Find the Slope of the Line

The slope (\( m \)) of the line is calculated by dividing the change in pressure by the change in temperature:\[m = \frac{\Delta P}{\Delta t} = \frac{37 \, \text{cm H}_2\text{O}}{10.8^{\circ} \text{C}} \approx 3.43 \, \text{cm H}_2\text{O}/^{\circ} \text{C}\]

Step 4: Solve for the Constant b

We use the equation \( P = mt + b \) with known values for one set of \( P \) and \( t \) to solve for \( b \). Using \( P = 1012 \, \text{cm H}_2\text{O} \) and \( t = 22.8^{\circ} \text{C} \):\[1012 = 3.43 \times 22.8 + b\]Simplifying the equation to find \( b \):\[1012 = 78.204 + b \b = 1012 - 78.204 = 933.796 \, \text{cm H}_2\text{O}\]

Step 5: Write the Full Equation

Substitute the values of \( m \) and \( b \) back into the equation \( P = mt + b \):\[ \boxed{P = 3.43t + 933.796} \]

Step 6: Find Temperature when Pressure is Zero

To find the temperature \( t \) when \( P = 0 \), substitute \( P = 0 \) into the equation and solve for \( t \):\[0 = 3.43t + 933.796 \]Rearrange and solve for \( t \):\[ 3.43t = -933.796 \t \approx -272.31^{\circ} \text{C}\]

Step 7: Transform to Kelvin Scale

To show the equation on the Kelvin scale where \( T = t + 272.31 \):\[ P = 3.43(t + 272.31) \]Expanding gives:\[P = 3.43T \]Thus, on the Kelvin scale, the equation simplifies to \( P = mT \).

Key Concepts

Kelvin Scale

The Kelvin scale is an absolute temperature scale used widely in scientific settings. It begins at absolute zero, which is the point where all molecular motion theoretically stops. Absolute zero is calculated to be approximately \(-273.15^{\circ} \text{C}\).

To convert a temperature from the Celsius scale to the Kelvin scale, you simply add \(273.15\) to the Celsius temperature. This means that when you have a formula like \(T = t + k\), \(T\) is the temperature in Kelvin and \(t\) is the temperature in Celsius. In the original exercise solution, this conversion is essential to simplify the temperature-pressure relationship and illustrates why Kelvin is used in scientific equations involving temperature.

Using Kelvin in formulas eliminates the problems associated with negative temperatures, allowing for a more straightforward relationship between temperature and other variables, like pressure.

Temperature-Pressure Relationship

In the context of gases, there's an interesting relationship between temperature and pressure that ties into Boyle's Law. Boyle's Law explains how gas pressure is inversely proportional to its volume, assuming temperature remains constant. However, when temperature changes, this direct relationship manifests: as the temperature of a gas increases in a closed container, its pressure increases as well, assuming the volume stays the same.

This relationship is demonstrated in the experiment from the exercise, where an increase in temperature leads to an increase in pressure. The calculated slope of the increase 'm' shows the rate of this change. Once the slope is established, you can express pressure as a linear function of temperature, \(P = mt + b\), demonstrating that pressure increases linearly with temperature. This linear relationship provides a simple reference for predicting pressure changes based on temperature variations.

Linear Expansion of Gases

When gases are heated, they expand. This concept, known as the linear expansion of gases, determines how gas volume or pressure changes with temperature. In the exercise, linear expansion is expressed as a mathematical equation \(P = mt + b\), where \(P\) is the pressure, \(m\) is the slope, and \(b\) is a constant.

This form of equation describes how, for small temperature changes, gases expand at a uniform rate, which is portrayed in the solution as the slope \(m\). This constant rate of expansion is a foundational aspect of gas behavior, making it essential to understand in thermodynamics and practical applications.

• Allows prediction of gas volume or pressure changes.
• Essential for designing equipment that involves gas under varying temperatures.
• Reflected in various natural and industrial processes, including weather patterns and engine mechanics.

The linear, predictable expansion illustrates how physical properties of gases adjust symmetrically with changes in temperature, reinforcing why equations like \(P = mt + b\) are widely applicable.