Problem 22 - University Physics with Modern Physics

Question

A cylinder 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 pV-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

(a) 4.989 J; (b) 8.315 J. More heat is needed at constant pressure due to expansion work.

Step 1: Understanding the Heat at Constant Volume

The heat required at constant volume for an ideal gas can be calculated using the formula: \[ Q_V = nC_V\Delta T \] where - \( Q_V \) is the heat added at constant volume, - \( n \) is the number of moles, - \( C_V \) is the molar heat capacity at constant volume, and - \( \Delta T \) is the change in temperature. For helium, a monoatomic ideal gas, \( C_V = \frac{3}{2} R \), where \( R \) is the universal gas constant \( 8.314 \text{ J/mol K} \). Calculate \( \Delta T = 67 - 27 = 40 \text{ K} \). Now substitute the known values: \[ Q_V = 0.0100 \times \frac{3}{2} \times 8.314 \times 40 \]

Step 2: Calculating Heat at Constant Volume

Using the values, the calculation for \( Q_V \) is as follows: \[ Q_V = 0.0100 \times \frac{3}{2} \times 8.314 \times 40 = 4.989 \text{ J} \].

Step 3: Understanding the Heat at Constant Pressure

When pressure is kept constant, the heat added \( Q_P \) is calculated by: \[ Q_P = nC_P\Delta T \], and for a monoatomic ideal gas \( C_P = \frac{5}{2} R \). Calculate \( \Delta T \) similarly as earlier, \( 40 \text{ K} \). Then calculate \( Q_P \): \[ Q_P = 0.0100 \times \frac{5}{2} \times 8.314 \times 40 \].

Step 4: Calculating Heat at Constant Pressure

Using the formula, calculate \( Q_P \) as: \[ Q_P = 0.0100 \times \frac{5}{2} \times 8.314 \times 40 = 8.315 \text{ J} \].

Step 5: Understanding the Difference in Heat

The difference between \( Q_P \) and \( Q_V \) is due to the additional work done by the gas when the volume is allowed to change under constant pressure. More heat is required at constant pressure because the gas expands and performs work against external pressure. The additional heat \( (Q_P - Q_V) \) accounts for this work.

Step 6: Comparing Internal Energy Changes in Cases (a) and (b)

For an ideal gas, the change in internal energy \( \Delta U \) is given by: \[ \Delta U = nC_V\Delta T \] and it is the same for both processes since internal energy change depends only on temperature change for an ideal gas. So, \( \Delta U = 4.989 \text{ J} \) in both cases, as calculated in Step 2.

Step 7: Summarizing pV Diagrams

For part (a), the pV diagram is a vertical line as volume remains constant. For part (b), the pV diagram is a horizontal line as pressure remains constant.

Key Concepts

Ideal Gas Law

The Ideal Gas Law combines several individual gas laws into one general equation. It describes the behavior of an ideal gas in terms of four variables: pressure \( P \), volume \( V \), temperature \( T \), and moles \( n \). The equation is represented as \( PV = nRT \), where \( R \) is a constant known as the universal gas constant (8.314 J/mol K). This equation helps us understand how changes in temperature, pressure, or volume affect a gas's behavior.
The Ideal Gas Law is essential in thermodynamics because it allows us to relate different states of a gas. It's particularly useful in processes where gases expand or contract, such as the isochoric or isobaric processes discussed in this problem. Ensuring that conditions are ideal is assumed, meaning no interactions between molecules other than perfectly elastic collisions, and that the molecules themselves occupy no volume.

• For the helium gas contained in the cylinder, this law helps predict how the gas will behave under different conditions, specifically when the temperature changes.

Heat Capacity

Heat capacity is a measure of how much heat energy a substance needs to change its temperature by a certain amount. It is specific as "molar heat capacity" for dealing with substances like our helium gas, meaning the heat capacity per mole of the substance.
There are two types of molar heat capacities relevant here:

• Constant Volume (\( C_V \)): For a monoatomic ideal gas like helium, the heat capacity at constant volume is \( C_V = \frac{3}{2}R \).
• Constant Pressure (\( C_P \)): For the same gas, the constant pressure heat capacity is \( C_P = \frac{5}{2}R \).

These values are derived from the degrees of freedom available to gas molecules and are critical in calculating the heat needed to change the temperature in different processes (like isochoric or isobaric).

Internal Energy

Internal energy is the total energy contained within a system, which for an ideal gas, comes solely from the kinetic energy of the gas molecules. It changes with temperature and is calculated using \( C_V \Delta T \).
For the helium gas in the cylinder, we calculated that the internal energy change \( \Delta U \) was 4.989 J when the temperature changed by 40 K in both isochoric and isobaric processes. This is because the change in internal energy for an ideal gas depends only on the temperature change, not on how the change is brought about (whether volume is constant or pressure is constant).

• This is a crucial aspect because it highlights the nature of energy conservation in thermodynamics.

Isobaric Process

An isobaric process is one where the pressure remains constant as the system undergoes a change such as heating or cooling. In such a process, work is done by the gas as it expands or contracts. This involves both the addition of heat and the performance of work.
For the helium gas, we calculated the heat needed under constant pressure as \( Q_P = 8.315 \text{ J} \). Since the pressure is constant but temperature is increasing, the gas expands, doing work on its surroundings. The heat added under these conditions does not only increase the internal energy but also accounts for the work done by the gas. As a result, more heat is required in this process than in an isochoric process with the same temperature change.

Isochoric Process

In an isochoric process, the volume remains constant while the system's temperature changes. Because the volume does not change, the system does no work. Any heat added directly changes the internal energy of the system.
For the exercise, the heat required at constant volume \( Q_V \) was found to be 4.989 J, which directly contributes to increasing the internal energy, with no part of it converting into work, unlike the isobaric process.

• The pV-diagram for an isochoric process shows a vertical line, indicating constant volume as pressure changes with temperature.

These characteristics make it simpler than the isobaric process but crucial for understanding how heat and energy interact in gas dynamics.