Problem 21 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

One mole of an ideal gas at \(300 \mathrm{~K}\) is expanded isothermally from an initial volume of \(1 \mathrm{~L}\) to \(10 \mathrm{~L}\). The change in internal energy, \(\Delta U\), for the gas in this process is (a) \(163.7 \mathrm{cal}\) (b) zero (c) \(1381.1 \mathrm{cal}\) (d) 9 L-atm

Step-by-Step Solution

Answer

(b) zero

Step 1: Understand the Process and the Gas Law

The problem describes an isothermal expansion of an ideal gas. In an isothermal process, the temperature remains constant. According to the first law of thermodynamics, \(\Delta U = Q - W\), where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added to the system, and \(W\) is the work done by the system. For an ideal gas undergoing an isothermal process, \(\Delta U = 0\) because the internal energy of an ideal gas is only a function of temperature, which stays constant in an isothermal process. Therefore, no change in internal energy occurs.

Step 2: Apply the First Law of Thermodynamics

According to the first law of thermodynamics, which states \(\Delta U = Q - W\), we can see that during an isothermal expansion, the internal energy does not change because the temperature is constant. The work done by the gas during expansion either equals the heat added to the system or comes from the internal energy of the gas. Since the internal energy of an ideal gas is solely dependent on temperature and the temperature is constant, the change in internal energy \(\Delta U\) is zero.

Step 3: Determine the Correct Answer for Change in Internal Energy

Given that the process is isothermal and the gas is ideal, we can conclude that the change in internal energy is zero. This corresponds to the options provided, and therefore, the correct answer is (b) zero.

Key Concepts

First Law of Thermodynamics

The first law of thermodynamics is a version of the law of conservation of energy, tailored for thermodynamic systems. It states that the energy added to a system in the form of heat, minus the energy lost as work done by the system, equals the change in internal energy of the system. Mathematically, it is expressed as \(\Delta U = Q - W\).

Understanding this law is crucial when analyzing processes involving ideal gases, as it helps to determine what will happen to the internal energy when heat and work are exchanged. Isothermal processes are a great example, as no matter how much work is done or heat is transferred, the internal energy remains constant because the temperature does not change. This is tightly correlated with the concept that for ideal gases, the internal energy is a function of temperature alone.

Internal Energy of Ideal Gas

In the context of an ideal gas, the internal energy is directly related to its temperature. This can be inferred from the kinetic molecular theory of gases, which posits that an ideal gas is composed of particles in constant, random motion. The energy of these particles, hence the internal energy of the gas, is dependent on their kinetic energy which is a function of temperature.

During an isothermal process, like the one given in the exercise, the temperature and thus the internal energy does not change. The formula for the internal energy of an ideal gas, \(U = \frac{3}{2}nRT\), where \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature, explains why the internal energy remains unchanged if the temperature remains constant. For the exercise's isothermal expansion, where temperature stays at 300 K, the internal energy change \(\Delta U\) is zero.

Work Done by Gas

When a gas expands, it does work on its surroundings. In thermodynamics, work done by a gas is an energy transfer from the gas to the surroundings. This is because to expand, the gas particles must exert a force over the distance they move against an external pressure.

In an isothermal process for an ideal gas, Henry Charles's Law helps calculate the work done. The work done by an isothermal expansion is given by \(W = nRT \ln(\frac{V_2}{V_1})\), where \(V_1\) and \(V_2\) are the initial and final volumes, respectively. Even though work is being done by the gas in the exercise (expanding from 1 L to 10 L), this energy transfer doesn't alter the internal energy. It does, however, reflect as heat added to the system (\(Q\)) to maintain constant temperature, demonstrating the interplay of work and heat in an isothermal process.

It's essential to understand that while work done by the gas might not change its internal energy in an isothermal process, it is a critical component of the system's energy exchange, and therefore of our understanding of how gases behave under different conditions.