Problem 20 - University Physics with Modern Physics

Question

Acylinder contains 0.0100 mol of helium at \(T=27.0^{\circ} \mathrm{C}\) . (a) How much heat is needed to raise the temperature to \(67.0^{\circ} \mathrm{C}\) while keeping the volume constant? Draw a \(p V\) -diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from \(27.0^{\circ} \mathrm{C}\) to \(67.0^{\circ} \mathrm{C} ?\) Draw a \(p V\) -diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?

Step-by-Step Solution

Answer

More heat is required for part b (8.32 J) than part a (4.99 J) due to work done by the gas. Internal energy change is the same for both, 4.99 J.

Step 1: Analyze the scenario

We have helium (monoatomic ideal gas) inside a cylinder. Given that there are 0.0100 mol of helium which initially have a temperature of 27.0°C.

Step 2: Calculate heat for constant volume (Part a)

For a process at constant volume, heat is given by:\[ Q = nC_V\Delta T \]where \( C_V = \frac{3}{2}R \) for monoatomic gases and \( \Delta T \) is the change in temperature from 27.0°C to 67.0°C. First, convert temperatures to Kelvin: 27.0°C = 300 K and 67.0°C = 340 K. Thus, \( \Delta T = 340 K - 300 K = 40 K \). Using \( R = 8.314 \, J/(mol\cdot K) \), the heat is: \[ Q = 0.0100 \times \frac{3}{2} \times 8.314 \times 40 = 4.99 \, J \].

Step 3: Plot pV Diagram for Part a

For a constant volume process, the pV diagram is a vertical line. As temperature increases at constant volume, pressure also increases on the vertical line.

Step 4: Calculate heat for constant pressure (Part b)

For constant pressure, heat is given by:\[ Q = nC_P\Delta T \] where \( C_P = \frac{5}{2}R \) for monoatomic gases. Using the same temperature change (\(\Delta T = 40 K\)): \[ Q = 0.0100 \times \frac{5}{2} \times 8.314 \times 40 = 8.32 \, J \].

Step 5: Plot pV Diagram for Part b

For a constant pressure process, the pV diagram is a horizontal line. As temperature increases at constant pressure, volume increases on the horizontal line.

Step 6: Determine the cause for heat difference (Part c)

The difference in the required heat is due to the work done by or on the gas during the process. In part (b) a constant pressure process, the gas does work on the surroundings as it expands, hence more heat is needed compared to a constant volume process (part a) where no work is done.

Step 7: Calculate internal energy change for Part a (Part d)

For an ideal gas, the change in internal energy \( \Delta U \) for constant volume is:\[ \Delta U = nC_V\Delta T \]Using values from Step 2:\[ \Delta U = 4.99 \, J \]

Step 8: Calculate internal energy change for Part b (Part d)

For an ideal gas, the change in internal energy \( \Delta U \) remains the same regardless of the process when considering only temperature changes:\[ \Delta U = 4.99 \, J \]

Step 9: Compare internal energy changes (Part d)

Since the internal energy change \( \Delta U \) only depends on the temperature change, both part (a) and part (b) have the same internal energy change of 4.99 J. This is because \( \Delta U \) for an ideal gas is path-independent and only depends on initial and final states.

Key Concepts

Constant Volume Process

In a constant volume process, as the name suggests, the volume of the gas is held constant while other properties such as temperature and pressure may change. When an ideal gas like helium is subjected to this process, the heat added to the gas, denoted by \(Q\), is used entirely to increase its internal energy. Since the volume doesn't change, no work is done by or on the gas. Therefore, the entire heat input goes into raising the temperature.

• The heat required for this process is calculated using the formula \( Q = nC_V\Delta T \), where \(n\) is the number of moles, \(C_V\) is the molar heat capacity at constant volume, and \(\Delta T\) is the temperature change.

For a monoatomic ideal gas like helium, \(C_V = \frac{3}{2}R\), where \(R\) is the ideal gas constant. In our example, the calculated heat input is 4.99 J.
The pressure can increase even though the volume stays constant, which results in a vertical line on a \(pV\) diagram. This line illustrates that as temperature rises, pressure increases as well.

Constant Pressure Process

A constant pressure process involves keeping the pressure of the gas unchanged while allowing volume changes as the temperature varies. In this process, the heat input not only increases the internal energy but also does work on expanding the gas.
This phenomenon is represented by using the equation \( Q = nC_P\Delta T \), where \(C_P = \frac{5}{2}R\) for monoatomic gases. Thus, the heat calculation in a constant pressure process results in a higher value because the added heat also accounts for the expansion work performed by the gas.
In the example provided, the calculated heat requirement is 8.32 J, showing more required energy compared to a constant volume process. The reason for this is that some of the heat goes into doing work while expanding. A \(pV\) diagram represents this process with a horizontal line, showing that as the gas is heated, the volume increases to maintain constant pressure.

Internal Energy Change

The internal energy change in an ideal gas depends solely on the change in temperature, regardless of the type of process it undergoes, as ideal gases are considered to be perfect in terms of behavior. This implies that the path taken to achieve this temperature change doesn't matter for calculating the change in internal energy \(\Delta U\).
The internal energy change can be calculated using the formula \( \Delta U = nC_V\Delta T \). In both constant volume and constant pressure processes from the example, the internal energy change remains the same at 4.99 J as long as the gas transitions between identical initial and final states in terms of temperature.
This invariability is what makes the internal energy change path-independent. Thus, while different processes require different amounts of heat, the change in internal energy is the same because it is a function of temperature change alone.

Ideal Gas Law

The ideal gas law is a key principle helping to understand the behavior of ideal gases, described by the equation \( PV = nRT \). This relationship correlates the pressure \(P\), volume \(V\), and temperature \(T\) of an ideal gas with its amount in moles \(n\) and the ideal gas constant \(R\).
In both constant volume and constant pressure scenarios, the ideal gas law aids in predicting how changing one state variable influences the others. For instance:

• In a constant volume process, increasing temperature leads to a rise in pressure.
• In a constant pressure process, when temperature rises, the volume must increase to maintain the same pressure.

Understanding this law allows us to graph behavior on \(pV\) diagrams and apply it to complex systems in thermodynamics. It's a fundamental tool in calculating and applying concepts such as heat input, internal energy change, and expansion work under varying conditions.