Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal ener

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Q. Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?

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Concept understanding — First Law of Thermodynamics

First Law of Thermodynamics

The Intuition: Energy is a Bank Account

Imagine you have a bank account. You can deposit money into it, or withdraw money from it. The total amount of money in your account changes only by the net of what goes in and what comes out. You cannot create money from nothing, nor can you destroy it — it just moves.

Energy works the same way. In any physical or chemical process, energy is never created or destroyed. It is only transferred from one place to another, or converted from one form to another. This is the First Law of Thermodynamics — the law of conservation of energy, applied to systems where heat and work are the currencies.

When you heat a gas in a piston, the gas expands and pushes the piston up. The energy you put in as heat doesn't vanish — part of it stays inside the gas (raising its temperature), and part of it leaves as work done on the piston. The total energy of the universe remains constant.

The Precise Statement

\Delta U = Q - W

Where:

$\Delta U$ = change in the internal energy of the system (the energy stored inside — kinetic energy of molecules, potential energy in bonds, etc.)
$Q$ = heat added to the system (positive if heat flows into the system)
$W$ = work done by the system on the surroundings (positive if the system expands and pushes against something)

The sign convention is crucial. Many textbooks use $Q + W$ with work done on the system. The version above ( $Q - W$ ) is the most common in Indian exam syllabi (CBSE, JEE, NEET). Stick to one convention and be consistent.

Watch out

Common Mistake

Students often forget the sign of work. If a gas expands, it does positive work on the surroundings — so $W$ is positive, and $\Delta U = Q - W$ becomes smaller. If a gas is compressed, work is done on the gas — so $W$ is negative, and $\Delta U = Q - (-|W|) = Q + |W|$ , which increases internal energy.

What Each Term Means Physically

Internal energy ( $U$ ) is the total microscopic energy of the system. For an ideal gas, it depends only on temperature — higher temperature means higher $U$ . For real substances, it also depends on volume and phase.

Heat ( $Q$ ) is energy transferred due to a temperature difference. If you put a hot pan on a cold stove, heat flows from pan to stove. In thermodynamics, we always ask: who is the system? If the system is the gas, then $Q$ is positive when heat flows into the gas.

Work ( $W$ ) in thermodynamics is usually pressure-volume work: $W = \int P \, dV$ . When a gas expands against a piston, it does work on the piston. When you compress a gas, you do work on it.

A Simple Example

Take a cylinder with a movable piston, containing 1 mole of an ideal gas. You supply 500 J of heat to the gas. The gas expands and does 200 J of work on the piston.

$Q = +500$ J (heat enters the system)
$W = +200$ J (work done by the system)

$\Delta U = 500 - 200 = 300 \text{ J}$ …

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Free expansion into a vacuum means no external work is done. For an ideal gas undergoing isothermal expansion, its internal energy remains unchanged. Thus, the work done is $\boxed{0}$ and the change in internal energy is $\boxed{0}$ .

When a gas expands, it does work on its surroundings, or the surroundings do work on the gas. The change in the internal energy of the gas is related to the heat exchanged and the work done, as described by the First Law of Thermodynamics. We need to understand what "free expansion" and "isothermal expansion" imply for an ideal gas.

Concept and Intuition

The First Law of Thermodynamics states that the change in the internal energy ( $\Delta U$ ) of a system is equal to the heat ( $Q$ ) added to the system minus the work ( $W$ ) done by the system on its surroundings.

\Delta U = Q - W

Let's break down the conditions given in the problem:

Free Expansion: This term describes an expansion where the gas expands against a vacuum. A vacuum implies that there is no external pressure ( $P_{ext} = 0$ ) opposing the expansion. If there's no external pressure, the gas does no work against the surroundings.
Isothermal Expansion: This means the temperature ( $T$ ) of the gas remains constant throughout the expansion.
Ideal Gas: For an ideal gas, the internal energy ( $U$ ) depends only on its temperature. It does not depend on its volume or pressure. This is a crucial property of ideal gases.

Combining these concepts, we can determine the work done and the change in internal energy.

Step-by-Step Calculation

Calculate the work done ( $W$ ) during free expansion. Work done by a gas expanding against an external pressure is generally given by $W = P_{ext} \Delta V$ , where $P_{ext}$ is the external pressure and $\Delta V$ is the change in volume. In free expansion, the gas expands into a vacuum, meaning the external pressure $P_{ext}$ is zero.

W = P_{ext} \Delta V

Since $P_{ext} = 0$:

W = (0) \times (V_{final} - V_{initial})

W = 0

Even though the volume changes from 1 litre to 5 litre, no work is done because there is no resistance to the expansion.

2. Calculate the change in internal energy ( $\Delta U$ ) for an ideal gas undergoing isothermal expansion.

The problem states that the gas is an ideal gas and the expansion is isothermal (constant temperature).

For an ideal gas, the internal energy $U$ is solely a function of its temperature $T$ .

$U = f(T)$ …

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Q.Expansion of a gas in vacuum is called free expansion. Calculate the work done and the change in internal energy when 1 litre of ideal gas expands isothermally into vacuum until its total volume is 5 litre?

Concept and Intuition

Step-by-Step Calculation

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