Problem 39 - University Physics with Modern Physics

Question

A quantity of air is taken from state \(a\) to state \(b\) along a path that is a straight line in the \(p V\) -diagram (Fig. Pl9.39). (a) In this process, does the temperature of the gas increase, decrease, or stay the same? Explain. (b) If \(V_{a}=0.0700 \mathrm{m}^{3}\) , \(V_{b}=0.1100 \mathrm{m}^{3}, \quad p_{a}=1.00 \mathrm{x}\) \(10^{5} \mathrm{Pa},\) and \(p_{b}=1.40 \times 10^{5}\) \(\mathrm{Pa},\) what is the work \(W\) done by that gas in this process? Assume that the gas may be treated as ideal.

Step-by-Step Solution

Answer

(a) The temperature changes. (b) The work done is 4800 J.

Step 1: Understanding Temperature Change

To assess if the temperature of the gas changes, note that the process is represented by a straight line on a pressure-volume \(pV\) diagram. For an ideal gas, temperature \(T\) is related to pressure \(p\), volume \(V\), and the number of moles \(n\) by the equation \(pV = nRT\). A straight line indicates a linear path for \(pV\), implying that the temperature changes unless the line represents an isothermal process. In this case, because both pressure and volume change, it indicates a change in temperature.

Step 2: Setting Up the Work Equation

To find the work done by the gas, recall that work done on or by the gas is the area under the process line on a \(pV\) diagram. For a process where the line is straight between two states \((a)\) and \((b)\), this is a trapezoidal area, calculated using the formula: \[W = \frac{1}{2} (p_a + p_b) (V_b - V_a)\].

Step 3: Calculating the Work Done

Input the given values into the work formula: \[W = \frac{1}{2} (1.00 \times 10^5 \mathrm{Pa} + 1.40 \times 10^5 \mathrm{Pa}) (0.1100 \mathrm{m}^3 - 0.0700 \mathrm{m}^3)\] Calculating this gives: \[W = \frac{1}{2} (2.40 \times 10^5 \mathrm{Pa}) (0.0400 \mathrm{m}^3)\] \[W = \frac{1}{2} (9.60 \times 10^3 \mathrm{J})\] \[W = 4.80 \times 10^3 \mathrm{J}\].

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in thermodynamics that describes the relationship between pressure, volume, and temperature in gases. It's expressed by the equation \(pV = nRT\), where:

• \(p\) is the pressure of the gas,
• \(V\) is the volume of the gas,
• \(n\) is the number of moles of the gas,
• \(R\) is the ideal gas constant, and
• \(T\) is the absolute temperature in Kelvin.

This equation assumes that the gas behaves ideally, meaning that it follows the assumptions of negligible intermolecular forces and volume. As pressure or volume changes, so must the temperature unless the number of moles of gas is also adjusted or the process is isothermal (occuring at a constant temperature).
In our exercise, as both pressure and volume change along the straight line in the \(pV\) diagram, the temperature must also change according to the Ideal Gas Law.

Work Done by Gas

Work done by a gas during expansion or compression can be visualized through the area under a curve in a \(pV\) diagram. Mathematically, work \(W\) for a process changing the state of the gas from (\(a\)) to (\(b\)) is given by:\[W = \int_{V_a}^{V_b} p \, dV\]This integration practically finds the area under the curve or line. For our particular exercise, where the process is represented by a straight line in the diagram between two points, this takes the form of a trapezoidal area. The formula:\[W = \frac{1}{2} (p_a + p_b) (V_b - V_a)\]helps in calculating the total work done by the gas. The work is positive if the gas expands (it does work on the surroundings), and negative if the gas is compressed (work is done on the gas).
Using the given values from our exercise, work done is calculated by substituting the pressure and volume at states \(a\) and \(b\).

pV Diagram

A \(pV\) Diagram (Pressure-Volume Diagram) is a graphical representation of the states of a gas and the work done during thermodynamic processes. It plots the pressure \(p\) on the y-axis against the volume \(V\) on the x-axis.
This visualization helps in determining how the gas behaves and the work done in processes such as compression and expansion. Common paths in a \(pV\) diagram include isothermal, isobaric, isochoric, and adiabatic processes.

• Isothermal Process: Constant temperature, curved paths.
• Isobaric Process: Constant pressure, horizontal lines.
• Isochoric Process: Constant volume, vertical lines.
• Adiabatic Process: No heat exchange, curved paths.

In the given exercise, the straight line connecting the two states indicates a linear relationship between pressure and volume as both change. It is indicative of a process that is neither of the primary thermodynamic processes listed above, meaning it's likely non-steady state, involving changes in internal energy and temperature.

Temperature Change in Gases

When dealing with gases, temperature is an essential factor influencing gas behavior in thermodynamic processes. Determining temperature change requires understanding how pressure and volume affect it. For an ideal gas, changing pressure and volume along a particular path will usually result in a change in temperature unless the process specifically maintains temperature constant such as in an isothermal process.

• Increase in Temperature: When either pressure or volume significantly increases while the other parameter doesn't offset.
• Decrease in Temperature: When either pressure or volume significantly decreases.
• Constant Temperature: If the increase of one parameter is exactly offset by the decrease of another at constant product \(pV\).

In the exercise, because both pressure and volume change along the straight line path in a \(pV\) diagram, the combined effects ensure a change in the temperature according to the Ideal Gas Law.
Understanding these changes helps in predicting how the gas will behave under different conditions and designing processes where temperature outcomes are vital.