Problem 68 - University Physics with Modern Physics

Question

Comparing Thermodynamic Processes. In a cylinder, 1.20 mol of an ideal monatomic gas, initially at \(3.60 \times 10^{5}\) Pa and \(300 \mathrm{K},\) expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a \(p V\) -diagram. In which case is the absolute value of the work done by the greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

Step-by-Step Solution

Answer

(a) Isothermal: 2740.4 J, (b) Adiabatic: -1376.4 J, (c) Isobaric: 5971.2 J; Greatest work in isobaric, least in adiabatic.

Step 1: Isothermal Expansion - Work Done

In isothermal processes, the temperature remains constant. The work done by an ideal gas during an isothermal expansion is given by the formula: \( W = nRT \ln \left( \frac{V_f}{V_i} \right) \), where \( n = 1.20 \text{ mol} \), \( R = 8.314 \text{ J/mol K} \), and \( T = 300 \text{ K} \). Since the volume triples, \( \frac{V_f}{V_i} = 3 \). Thus,\[W = 1.20 \cdot 8.314 \cdot 300 \cdot \ln(3) \approx 2740.4 \text{ J}.\]

Step 2: Adiabatic Expansion - Work Done

For an adiabatic process where no heat exchange occurs \((Q = 0)\), the work done can be calculated using \( W = \frac{(P_iV_i^{\gamma} - P_fV_f^{\gamma})}{\gamma - 1} \). The value of \( \gamma \) for a monatomic gas is \( \frac{5}{3} \). First, determine \( V_i \) from \( PV = nRT \):\[ V_i = \frac{nRT}{P_i} = \frac{1.20 \times 8.314 \times 300}{3.60 \times 10^5} = 0.00832 \text{ m}^3. \]Now, \( V_f = 3V_i = 0.02496 \text{ m}^3 \). Using the adiabatic condition \( P_iV_i^\gamma = P_fV_f^\gamma \), find \( P_f \): \( P_f = 3.60 \times 10^5 \left( \frac{1}{3} \right)^\gamma \). Substitute \( P_f \) into the work equation (you calculated \( P_f \) previously) and solve for \( W \): \[W = \frac{3.60 \times 10^5 \times (0.00832^{5/3} - 0.02496^{5/3})}{\frac{5}{3} - 1}.\] Calculate this to find \( W \approx -1376.4 \text{ J} \).

Step 3: Isobaric Expansion - Work Done

In an isobaric process, the pressure remains constant. The work done by the gas can be calculated as \( W = P(V_f - V_i) \). We know \( P = 3.60 \times 10^5 \text{ Pa} \), \( V_i = 0.00832 \text{ m}^3 \), and \( V_f = 3 \times V_i = 0.02496 \text{ m}^3 \). Thus,\[W = 3.60 \times 10^5 \times (0.02496 - 0.00832) \approx 5971.2 \text{ J}.\]

Step 4: Plot the Processes on a PV-Diagram

To visualize the processes, plot the initial state \((p_i,V_i)\) and the final state \((p_f,V_f)\) on a PV-diagram. The isothermal process will be a hyperbolic curve, the adiabatic process will be steeper than the isothermal curve, and the isobaric process will be a horizontal line indicating constant pressure as volume increases.

Step 5: Determine the Process with Greatest and Least Work Done

From the calculations:- Isothermal work done: \( 2740.4 \text{ J} \)- Adiabatic work done: \( -1376.4 \text{ J} \)- Isobaric work done: \( 5971.2 \text{ J} \)The greatest absolute work is in the isobaric process, and the least is in the adiabatic process.

Step 6: Heat Transfer in Each Process

In the isothermal process, \( Q = W \approx 2740.4 \text{ J} \) because temperature remains constant. In the adiabatic process, \( Q = 0 \text{ J} \) because no heat is exchanged. In the isobaric process, use \( Q = nC_p \Delta T \). \( C_p = \frac{5R}{2} \) for monatomic gases, and \( \Delta T \approx \frac{3W}{5R} \). Calculate the specific values to determine each \( Q \). The greatest absolute value of heat transfer is in the isobaric process, and the least in the adiabatic process.

Step 7: Change in Internal Energy for Each Process

According to the first law of thermodynamics, \( \Delta U = Q - W \). For the isothermal process, \( \Delta U = 0 \), as it's only a function of temperature. For the adiabatic process, \( \Delta U = -W \) since \( Q = 0 \). For the isobaric process, calculate \( \Delta U = Q - W \) using the values from previous steps. The greatest absolute internal energy change is in the adiabatic process and the least in the isothermal process.

Key Concepts

Isothermal Expansion

In an isothermal expansion, the temperature of the system remains constant, which is crucial as it dictates that the internal energy does not change. For an ideal gas undergoing isothermal expansion, the formula to calculate work done, \( W \), is given by:

• \( W = nRT \ln \left( \frac{V_f}{V_i} \right) \)

Here:

• \( n \) is the number of moles of the gas
• \( R \) is the ideal gas constant
• \( T \) is the absolute temperature (constant in this case)
• \( V_f \) and \( V_i \) are the final and initial volumes respectively

Specific to the example, since the volume triples, the work done is calculated to be approximately 2740.4 Joules. The key takeaway from isothermal processes is the importance of maintaining constant temperature, which ensures that heat flows into or out of the system as the gas does work, keeping internal energy unchanged.

Adiabatic Process

In an adiabatic process, the system is perfectly insulated, which means no heat exchange occurs with the surroundings. This is marked by \( Q = 0 \), and any work done results solely in the change in internal energy of the gas.The work done in an adiabatic process can be calculated using the equation:\[ W = \frac{(P_i V_i^{\gamma} - P_f V_f^{\gamma})}{\gamma - 1} \]For monatomic gases like the one in our exercise, the adiabatic index \( \gamma \) is \( \frac{5}{3} \).The challenge here is to understand the balance between pressure, volume, and internal energy without external heat exchange. In the provided example, the work done by the gas calculated with these principles comes to approximately \(-1376.4\) Joules. Since there is no heat exchange, the energy involved in expansion manifests solely as work done by or on the gas, making adiabatic processes distinctively informed by its inherent mechanical properties.

Isobaric Expansion

Isobaric processes occur at constant pressure. In such processes, the work done by the gas can be simply calculated using:

• \( W = P(V_f - V_i) \)

Where:

• \( P \) is the constant external pressure
• \( V_f \) and \( V_i \) are, once again, the final and initial volumes

This setup simplifies the process to straightforward volume calculations if pressure remains constant, as in our example. The result is a significantly large value of work, approximately 5971.2 Joules, highlighting the efficiency of isobaric processes when volume changes occur at steady pressure levels. It is important to note that since pressure is constant, any added heat will translate directly into work and a change in the internal energy of the gas.

Ideal Gas Law

The Ideal Gas Law is a fundamental principle that relates pressure, volume, and temperature of an ideal gas. Expressed by the equation:\( PV = nRT \)Where each variable represents:

• \( P \) - pressure of the gas
• \( V \) - volume it occupies
• \( n \) - number of moles of gas present
• \( R \) - universal gas constant
• \( T \) - absolute temperature

This law is crucial across all thermodynamic processes and was effectively used to establish initial conditions in the problem for calculating work done in each distinct scenario. The Ideal Gas Law makes it clear that altering any single variable will necessitate a change in another for the equation to maintain its truth, providing a foundational understanding of how ideal gases behave under various conditions.

Work Done by Gas

Work done by a gas during expansion or compression is a vital concept in thermodynamics. It illustrates how energy transfer occurs in physical systems when volume changes. Work \( W \) in thermodynamics is generally associated with the equation:

• \( W = P \Delta V \)

In processes such as those examined — isothermal, adiabatic, and isobaric — work calculations reveal how gases perform and transfer energy when external conditions like pressure and heat exchange change.In our example:

• Isothermal work: 2740.4 Joules
• Adiabatic work: \(-1376.4\) Joules
• Isobaric work: 5971.2 Joules

Understanding these calculations gives insight into how different forces act on and within a gas, ultimately affecting the change in volume and the energy dynamics of the system.