Question

An insulated container is divided into two equal portions. One portion contains one mole of an ideal monoatomic gas at pressure \(\mathrm{P}\) and temperature \(\mathrm{T}\), while the other portion is a perfect vacuum. If the intermediate partition is removed and the gas is allowed to expand, the change in the internal energy of gas is (1) \(\frac{3}{2} \mathrm{RT}\) (2) \(\mathrm{PV}\) (3) \(\frac{\mathrm{RT}}{2}\) (4) Zero

Short answer

The change in internal energy is zero.

Step-by-step solution

Understanding the Problem

There is an insulated container with two equal portions. One portion contains one mole of a monoatomic ideal gas at pressure \(\mathrm{P}\) and temperature \(\mathrm{T}\), and the other portion is a vacuum. When the partition is removed, the gas expands.

First Law of Thermodynamics

According to the first law of thermodynamics, the change in internal energy \(\Delta \mathrm{U}\) of a system is given by \(\Delta \mathrm{U} = q - w\), where \(q\) is the heat added to the system and \(w\) is the work done by the system.

Expansion into Vacuum

When the gas expands into a vacuum (free expansion), there is no external pressure, hence no work is done by the gas. Therefore, \(w = 0\).

Insulated Container

Since the container is insulated, there is no heat exchange with the surroundings, so \(q = 0\).

Change in Internal Energy

With \(q = 0\) and \(w = 0\), the change in internal energy \(\Delta \mathrm{U}\) is \(\Delta \mathrm{U} = 0\).

Key concepts

first law of thermodynamics

The first law of thermodynamics is a fundamental principle that relates the internal energy change in a system to heat added and work done by the system. It can be stated as: \ \( \Delta U = q - w \) \ Where:

• \( \Delta U \) is the change in internal energy
• \( q \) is the heat added to the system
• \( w \) is the work done by the system

In simpler terms: This law ensures energy conservation. Energy cannot be created or destroyed, only transferred or converted in form. For example, if you heat a gas, its internal energy increases, but if the gas does work (like pushing a piston), some of that energy is used up. Understanding this law is essential for analyzing thermodynamic processes and determining how energy changes within a system.

internal energy change

Internal energy is the total energy contained within a system, arising from both the kinetic and potential energies of its particles. In the context of an ideal gas, kinetic energy is especially important. For a monoatomic ideal gas, internal energy depends solely on temperature. We express this relation as: \( U = \frac{3}{2}nRT \), where:

• \( U \) is internal energy
• \( n \) is the number of moles
• \( R \) is the gas constant
• \( T \) is the temperature

In the given exercise, if no heat is added (\( q = 0 \)) and no work is done (\( w = 0 \)), the change in the internal energy ( \( \Delta U \)) is zero. So, the gas expands into a vacuum without changing its internal energy, maintaining the same temperature and internal energy as before expansion.

insulated container

An insulated container is a system that does not exchange heat with its surroundings. This means the container keeps the heat within its boundaries. Key features of an insulated container:

• No heat transfer (\( q = 0 \))
• The work done by expanding gas is against a vacuum, with no external pressure\( w = 0 \)

In the given problem, since the container is insulated, the conditions ensure no heat input or output. Thus, the first law of thermodynamics simplifies to: \( \Delta U = q - w = 0 \) Therefore, when the gas expands in this insulated environment, its internal energy remains unchanged.