Problem 2
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 \(30.9^{\circ} \mathrm{C}\), the pressure of the air went from \(1021\space \mathrm{cm}\space \mathrm{H}_{2} \mathrm{O}\) up to \(1049 \space\mathrm{cm} \space\mathrm{H}_{2} \mathrm{O}\). Since the air expands linearly with temperature, the equation relating \(P\) to \(t\) is of the form: $$P=m t+b$$ 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 Eq. 9 and solve for \(b\).) \(b=\)___________ \(\mathrm{cm} \space\mathrm{H}_{2} \mathrm{O}\)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 \(t=\)_____________\(^{\circ} \mathrm{C}=t_{0}=-k\) (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 K. On that scale, \(T=t+k\). Show that, on the Kelvin scale. your equation reduces to \(P=m T\).
Step-by-Step Solution
Verified Answer
a. m = 3.46 cm H₂O/°C
b. b = 942.11 cm H₂O
c. P = 3.46t + 942.11
P = 0 at t ≈ -272.17°C
e. On Kelvin scale: P = 3.46T
1Step 1: Calculate the change in pressure (ΔP)
To find the change in pressure, subtract the initial pressure from the final pressure:\[\Delta P = 1049 \ \text{cm H}_2\text{O} - 1021 \ \text{cm H}_2\text{O} = 28 \ \text{cm H}_2\text{O}\]
2Step 2: Calculate the change in temperature (Δt)
To find the change in temperature, subtract the initial temperature from the final temperature:\[\Delta t = 30.9^{\circ} \text{C} - 22.8^{\circ} \text{C} = 8.1^{\circ} \text{C}\]
3Step 3: Find the slope (m) of the line
The slope can be calculated as the change in pressure divided by the change in temperature:\[m = \frac{\Delta P}{\Delta t} = \frac{28 \ \text{cm H}_2\text{O}}{8.1^{\circ} \text{C}} \approx 3.46 \ \text{cm H}_2\text{O}/^{\circ} \text{C}\]
4Step 4: Solve for the y-intercept (b)
Substitute a known pair of values for \(P\) and \(t\) into the equation \(P=mt+b\) and solve for \(b\). Using the initial conditions \(P = 1021\) cm H\(_2\)O at \(t = 22.8^{\circ} \text{C}\): \[1021 = 3.46 \times 22.8 + b\]\[1021 = 78.888 + b\]\[b = 1021 - 78.888 \approx 942.11 \ \text{cm H}_2\text{O}\]
5Step 5: Write the equation with known m and b values
Substitute the calculated \(m\) and \(b\) into the equation \(P=mt+b\): \[P = 3.46t + 942.11\]
6Step 6: Solve for the temperature when P = 0
Set \(P = 0\) and solve for \(t\): \[0 = 3.46t + 942.11\]\[3.46t = -942.11\]\[t \approx \frac{-942.11}{3.46} \approx -272.17^{\circ} \text{C}\]
7Step 7: Express the equation using Kelvin scale
On the Kelvin scale, \(T = t + 272.17\). Substitute \(t = T - 272.17\) into the equation: \[P = 3.46(T - 272.17) + 942.11\] \[P = 3.46T - 3.46 \times 272.17 + 942.11\]Simplifying, the intercept \(-3.46 \times 272.17 + 942.11\) becomes zero. Thus, the equation reduces to: \[P = 3.46T\]
Key Concepts
Temperature-Pressure RelationshipLinear ExpansionKelvin ScaleAbsolute Zero
Temperature-Pressure Relationship
The relationship between temperature and pressure of a gas is an important concept in physics and chemistry. As the temperature of a gas increases, the particles within it move more energetically. This increased movement results in more frequent and forceful collisions with the walls of the container, which in turn increases the pressure exerted by the gas. Conversely, when temperature decreases, the movement of gas particles slows down, leading to fewer collisions and lower pressure.
This concept is fundamental to understanding how gases behave under various conditions. It is captured quantitatively by Charles's Law and Gay-Lussac's Law. In this exercise, the formula \[ P = m t + b \] illustrates a linear relationship where:
This concept is fundamental to understanding how gases behave under various conditions. It is captured quantitatively by Charles's Law and Gay-Lussac's Law. In this exercise, the formula \[ P = m t + b \] illustrates a linear relationship where:
- P is the pressure of the gas.
- m is the slope, representing the rate of change of pressure with temperature.
- t is the temperature in Celsius.
- b is a constant related to the initial state of the gas.
Linear Expansion
Linear expansion refers to the change in volume or size of a substance when its temperature changes. This is particularly relevant for gases, where an increase in temperature generally results in an increase in volume or pressure, assuming the size of the container remains constant.
In a system where gas expansion is linear, as assumed in the exercise, the change in pressure is directly proportional to the change in temperature. This can be calculated using the slope formula:\[ m = \frac{\Delta P}{\Delta t} \] Linear expansion is crucial for many practical applications:
In a system where gas expansion is linear, as assumed in the exercise, the change in pressure is directly proportional to the change in temperature. This can be calculated using the slope formula:\[ m = \frac{\Delta P}{\Delta t} \] Linear expansion is crucial for many practical applications:
- It helps engineers design systems that manage temperature changes efficiently.
- It plays a role in atmospheric science for predicting how the atmosphere will react to temperature changes.
- It is used in industrial applications where temperature control affects pressure and volume.
Kelvin Scale
The Kelvin scale is an absolute temperature scale widely used in scientific measurements. It is based on absolute zero, the theoretical point where all molecular motion ceases, defined as 0 Kelvin (K). Unlike the Celsius and Fahrenheit scales, Kelvin starts at the lowest possible temperature physicists believe the universe can reach.
When working with gas laws and other thermodynamic principles, using the Kelvin scale simplifies calculations and provides meaningful results because it directly correlates to the actual kinetic energy of the particles. The relationship between Celsius and Kelvin is straightforward:\[ T(K) = t(^{\circ}C) + 273.15 \]This conversion is essential for ensuring that temperature-related calculations are consistent and accurate.
When working with gas laws and other thermodynamic principles, using the Kelvin scale simplifies calculations and provides meaningful results because it directly correlates to the actual kinetic energy of the particles. The relationship between Celsius and Kelvin is straightforward:\[ T(K) = t(^{\circ}C) + 273.15 \]This conversion is essential for ensuring that temperature-related calculations are consistent and accurate.
- In practical applications, using Kelvin prevents confusion associated with negative temperatures experienced in Celsius or Fahrenheit scales.
- It is crucial for physics and chemistry where precision and harmony in data are needed.
- Kelvin is integral to theoretical predictions concerning temperature phenomena.
Absolute Zero
Absolute zero is a fundamental concept in physics. It is theoretically the lowest possible temperature, defined as 0 K (Kelvin) or -273.15°C (Celsius). At this temperature, particles have minimal vibrational motion, as zero kinetic energy in an ideal state is theoretically unattainable.
Reaching absolute zero is impossible; however, understanding the concept helps scientists comprehend the behavior of matter at extremely low temperatures. It stands as a cornerstone of thermodynamics and is central to the derivation and understanding of the Kelvin scale.
In experimental settings, the pursuit of reaching temperatures close to absolute zero has allowed humans to discover fascinating phases of matter, such as Bose-Einstein Condensates, where particles behave in unusual ways. Experimentally, while achieving absolute zero remains out of reach, it:
Reaching absolute zero is impossible; however, understanding the concept helps scientists comprehend the behavior of matter at extremely low temperatures. It stands as a cornerstone of thermodynamics and is central to the derivation and understanding of the Kelvin scale.
In experimental settings, the pursuit of reaching temperatures close to absolute zero has allowed humans to discover fascinating phases of matter, such as Bose-Einstein Condensates, where particles behave in unusual ways. Experimentally, while achieving absolute zero remains out of reach, it:
- Provides insight into the thermal properties of materials.
- Helps refine laws regarding energy transfer and entropy.
- Offers guidance in developing technologies for extreme cooling and improving superconductivity.