Question

In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60 \(\times\) 10\(^5\) Pa and 300 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 \(pV\)-diagram. In which case is the absolute value of the work done by the gas 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?

Short answer

Greatest work: Isothermal; Least work: Adiabatic. Greatest heat transfer: Isothermal; Least: Adiabatic. Greatest internal energy change: Adiabatic; Least: Isothermal.

Step-by-step solution

Identify Given Information

We have 1.20 mol of an ideal monatomic gas initially at pressure \( P_1 = 3.60 \times 10^5 \) Pa and temperature \( T_1 = 300 \) K. The gas expands to triple its initial volume. We need to calculate the work done for three processes: isothermal, adiabatic, and isobaric.

Step 2(a): Calculate Work for Isothermal Expansion

In an isothermal process, the work done by the gas is given by \( W = nRT \ln\left( \frac{V_2}{V_1} \right) \). Since the volume triples, \( \frac{V_2}{V_1} = 3 \). Using the ideal gas constant \( R = 8.314 \) J/mol·K, and \( n = 1.20 \), \( T = 300 \), substitute these values:\[ W = 1.20 \times 8.314 \times 300 \times \ln(3) \]Calculate \( W \approx 2.73 \times 10^3 \text{ J} \).

Step 2(b): Calculate Work for Adiabatic Expansion

In an adiabatic process, the relationship \( PV^\gamma = \text{constant} \) holds, where \( \gamma = \frac{5}{3} \) for a monatomic ideal gas. Use \( \frac{V_2}{V_1} = 3 \) and the initial conditions to find work:\[ W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} \].First, find \( P_2 \) using \( P_1V_1^\gamma = P_2V_2^\gamma \) then substitute to find the work. This involves solving:\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^\gamma = 3.60 \times 10^5 \times \left( \frac{1}{3} \right)^{5/3} \].Calculate the work \( W \approx 1.60 \times 10^3 \text{ J} \).

Step 2(c): Calculate Work for Isobaric Expansion

In an isobaric process, the pressure remains constant. The work done is given by \( W = P \Delta V \). Since the volume triples, \( \Delta V = 2V_1 \). From the ideal gas law, \( V_1 = \frac{nRT}{P} \). Substitute \( P = 3.60 \times 10^5 \),\[ V_1 = \frac{1.20 \times 8.314 \times 300}{3.60 \times 10^5} \].Then, \( W = 3.60 \times 10^5 \times 2 \times V_1 \). Calculate \( W \approx 1.99 \times 10^3 \text{ J} \).

Compare Work Done

Compare the calculated work done in each process:- Isothermal: \( W \approx 2.73 \times 10^3 \) J (greatest)- Isobaric: \( W \approx 1.99 \times 10^3 \) J- Adiabatic: \( W \approx 1.60 \times 10^3 \) J (least)The greatest work done is during the isothermal process.

Compare Heat Transfer and Change in Internal Energy

For heat transfer: - In an isothermal process, heat transfer is maximized. - In an adiabatic process, there is no heat transfer. For change in internal energy: - In an isothermal process, internal energy does not change. - In an adiabatic process, the internal energy change is largest compared to isobaric and minimal in isothermal.

Key concepts

Isothermal Expansion

In an isothermal expansion, the temperature of an ideal gas remains constant throughout the process. This implies that even as the gas expands and does work on its surroundings, it must absorb heat from the environment to maintain a constant temperature. The work done by the gas can be calculated using the formula:

• \( W = nRT \ln\left( \frac{V_2}{V_1} \right) \)

Here, \( n \) is the amount of substance in moles, \( R \) is the ideal gas constant, \( T \) is the temperature in Kelvin, and \( V_2/V_1 \) is the volume ratio. During an isothermal process, because there is no change in internal energy, the heat added to the system equals the work done by the gas.
This makes isothermal processes unique as it means the system exchanges the most heat with its surroundings among all types of thermodynamic processes.

Adiabatic Expansion

An adiabatic expansion occurs when gas expands without exchanging heat with its surroundings. During this process, all the work done by the gas causes a drop in its internal energy, resulting in a decrease in temperature. The adiabatic condition follows the relationship:

• \( PV^\gamma = \text{constant} \)

where \( \gamma = \frac{C_p}{C_v} \) is the heat capacity ratio. For an ideal monatomic gas, \( \gamma \) is 5/3.
Adiabatic processes can be fast because there's no concern about the heat exchange.
This process shows the greatest change in temperature and internal energy, balancing the reduced work done compared to isothermal expansion.

Isobaric Process

An isobaric process occurs under constant pressure. In this scenario, when a gas expands, the work done is simply the product of the constant pressure and the change in volume. This can be expressed as:

• \( W = P \Delta V \)

where \( \Delta V \) is the change in volume, calculated as the difference between final and initial volumes. This type of process is frequently encountered in everyday scenarios, such as heating air in a piston at constant pressure.
Though the amount of work done by the gas is more than in an adiabatic expansion, it is still less than the work done during an isothermal expansion when all other factors remain constant.

Work Done by Gas

Work done by gas in any thermodynamic process can be understood as the energy transferred by the system to its surroundings due to changes in volume. For ideal gases, understanding how different processes affect work done is crucial:

• Isothermal: Maximum work done as pressure decreases slowly over a large distance because temperature remains constant.
• Adiabatic: Least work done because of no heat exchange and subsequent rapid volume change.
• Isobaric: Intermediate work done, as pressure stays the same throughout the volume change.

The concept of work done is key in determining the efficiency of mechanical engines, refrigeration cycles, and various other applications in thermodynamics.

Heat Transfer Calculations

Heat transfer is a crucial concept that defines how energy moves in and out of a gas during thermodynamic processes. It varies with each type of process:

• Isothermal: Heat transfer is non-zero and equals the work done, making it substantial. This is because temperature remains constant and heat compensates for the work done by the gas.
• Adiabatic: There is no heat transfer. All energy changes occur internally through work done by or on the gas.
• Isobaric: The heat transfer depends on both work done and changes in internal energy. Due to constant pressure, heat must be added or removed to change temperature.

Understanding heat transfer helps in designing systems where temperature control is essential, such as engines, and maintaining energy efficiency.