Question

In this experiment a student found that when she increased the temperature of a \(550 \mathrm{mL}\) sample of air from \(22.3^{\circ} \mathrm{C}\) to \(29.7^{\circ} \mathrm{C},\) the pressure of the air went from \(1009 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\) up to \(1033 \mathrm{cm} \mathrm{H}_{2} \mathrm{O}\). Since the air expands linearly with temperature, the equation relating \(P\) to \(t\) is of the form: $$P=m t+b$$ 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 \(b\).)\(b=\)_____ c. Express Equation 9 in terms of the values of \(m\) and \(b\). \(P=\) d. At what temperature \(t\) will \(P\) become zero? \(P=0 \quad\) at \(\quad t=\)_____ \(^{\circ} \mathrm{C}=t_{\mathrm{o}}=-A\) 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, \(T=t+A\). Show that, on the Kelvin scale, your equation reduces to \(P=m T\).

Short answer

Solution: Step 1: Calculate the slope (m) \(m = \frac{ΔP}{Δt} = \frac{1033 - 1009}{29.7 - 22.3} = \frac{24}{7.4} = 3.24\; cm \; H_{2}O/ ^{\circ} C\) Step 2: Find the value of 'b' \(b = P1 - m * t1 = 1009 - 3.24 * 22.3 = 1009 - 73.12 = 935.88\) Step 3: Express the equation for P in terms of m and b \(P = 3.24t + 935.88\) Step 4: Determine the temperature at which P becomes zero \(t = -\frac{b}{m} = -\frac{935.88}{3.24} = -288.79^{\circ} C\) Step 5: Convert the equation to Kelvin scale \(P = mT\)

Step-by-step solution

Calculate the slope (m)

We are asked to find the slope (m) of the line, which represents the change in pressure (P) divided by the change in temperature (t). We can calculate the slope using the formula: \(m = \frac{ΔP}{Δt}\) where, ΔP = P2 - P1 and Δt = t2 - t1 Given, P1 = \(1009\; cm\; H_{2}O\) P2 = \(1033\; cm\; H_{2}O\) t1 = \(22.3^{\circ} C\) t2 = \(29.7^{\circ} C\) Let us now calculate the slope (m).

Find the value of 'b'

To find the value of 'b', we can substitute any pair of P and t into the equation P = mt + b and solve for 'b'. Let's use the initial values P1 and t1: \(P1 = m * t1 + b\) Substituting the values for P1, m, and t1, and solving for b: \(b = P1 - m * t1\)

Express the equation for P in terms of m and b

Now that we have calculated the values for m and b, we can substitute them back into the original equation for P: \(P = mt + b\)

Determine the temperature at which P becomes zero

We now need to find the temperature (t) at which the pressure (P) becomes zero: \(P = mt + b\) \(0 = mt + b\) To find the temperature (t), we can rearrange the equation and solve for t: \(t = -\frac{b}{m}\)

Convert the equation to Kelvin scale

The relationship between Celsius and Kelvin scales is simple: \(T = t + A\) where T is the temperature in Kelvin, t is the temperature in Celsius, and A is the absolute zero of temperature in Celsius. From the previous step, we found that: \(t_o = -A\) Now, replacing t with T - A in the equation for P: \(P = m(T - A) + b\) Since \(t_o = -A\), and when \(t = t_o\), \(P = 0\): \(0 = m(t_o) + b\) Now, substituting this into the previous equation: \(P = m(T - A) - m(t_o)\) So, the equation in terms of Kelvin temperature scale becomes: \(P = mT\)

Key concepts

Pressure-Temperature Relationship

The pressure-temperature relationship describes how the pressure of an ideal gas changes with temperature. This concept is based on Gay-Lussac’s Law, which states that the pressure of a gas at constant volume is directly proportional to its absolute temperature. In the context of this experiment, as the temperature of the air increases from 22.3°C to 29.7°C, the pressure also increases.

Here's how it works:

• The initial pressure, \( P_1 = 1009 \) cm H extsubscript{2}O, corresponds to the initial temperature, \( t_1 = 22.3^{ ext{o}} ext{C} \).
• As the temperature increases to \( t_2 = 29.7^{ ext{o}} ext{C} \), the pressure rises to \( P_2 = 1033 \text{ cm H}_2 ext{O} \).

This data helps in understanding how the changes in temperature cause proportional changes in pressure at a constant volume. In general, if the temperature of a gas increases while the volume remains constant, the gas molecules move more energetically and collide with the walls of their container more frequently, thus increasing the pressure.

Absolute Zero

Absolute zero is a theoretical temperature point where the motion of particles theoretically ceases, implying zero kinetic energy in gases. This concept is vital because it forms the foundation of the Kelvin temperature scale, famously introduced by Lord Kelvin. In the exercise, when pressure \( P \) becomes zero, it theoretically suggests that the temperature has reached absolute zero.

Steps involved in calculating absolute zero:

• The equation derived from the pressure-temperature relationship is \( P = mt + b \).
• For \( P = 0 \), solve for temperature \( t \) using \( 0 = mt + b \), leading to \( t = -\frac{b}{m} \).

This temperature, \( t \), represents absolute zero when converted to Kelvin. It is essential to realize that absolute zero is unachievable practically but serves as a critical boundary in thermodynamics and carnets our understanding of particle behavior at low temperatures.

To adopt this in the Kelvin scale:

• Absolute zero in Celsius (°C) is approximately -273.15°C.
• On the Kelvin scale, this corresponds to 0 K.

This transformation aligns scientific measurements and simplifies the calculation of thermodynamic equations by eliminating negative values.

Linear Expansion of Gases

The linear expansion of gases principle states that gas volume expands linearly with an increase in temperature if the pressure is kept constant. However, the reverse interpretation is observed in this experiment: pressure grows with temperature under constant volume.

Key points about linear expansion of gases:

• In a sealed container where volume cannot change, increasing temperature forces pressure to increase instead.
• The experiment provided values support the linear relationship via the equation \( P = mt + b \).
• The known values of temperature and pressure can be used to derive the slope \( m \) of this line, showing how much pressure changes per unit temperature increase.

The insight gained from understanding linear gas expansion is pivotal in many fields, such as meteorology and automotive engineering, where control of gas expansion under varying environmental conditions is vital. Recognizing linear behaviors aids in predicting outcomes and allows for engineering designs that accommodate thermal changes, ensuring the reliability and safety of systems that depend on gas pressure.