A gas is compressed inside a cylinder, and 462 J of work is done on the gas. During the same process, the gas releases 128 J of heat to the surroundings.Using the first law of thermodynamics, calculate the change in the internal energy of the gas.

A gas is compressed inside a cylinder, and 462 J of work is done on the gas. During the same process, the gas releases 128 J of heat to the surroundings.

Using the first law of thermodynamics, calculate the change in the internal energy of the gas.

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Interpretation

This problem is based on the First Law of Thermodynamics, which relates heat, work, and the change in the internal energy of a system.

The governing equation is:

\(\Delta E = q + w\)

where:

• \(\Delta E\) = change in internal energy of the system
• \(q\) = heat exchanged with the surroundings
• \(w\) = work done on or by the system

The most important part of such questions is assigning the correct signs to \(q\) and \(w\).

• If the system absorbs heat, then \(q > 0\).
• If the system releases heat, then \(q < 0\).
• If work is done on the system (compression), then \(w > 0\).
• If the system does work on the surroundings (expansion), then \(w < 0\).

A common mistake is to ignore these sign conventions. The calculation itself is simple once the signs are chosen correctly.

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Concept Application

Here, the system is the gas.

The gas is compressed, meaning the surroundings push on the gas. Since work is done on the system,

\(w = +462\ \text{J}\)

The gas also releases heat to the surroundings. Heat leaving the system means

\(q = -128\ \text{J}\)

Now both quantities are ready to substitute into the First Law.

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Solution

Using the First Law of Thermodynamics,

\(\Delta E = q + w\)

Substitute the given values:

\(\Delta E = (-128\ \text{J}) + (462\ \text{J})\) \(\Delta E = 334\ \text{J}\)

Therefore,

The internal energy of the gas increases by \(+334\ \text{J}\).

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Insight

A simple way to remember the sign convention is:

• Compression → Work done on the system → Positive \(w\)
• Expansion → Work done by the system → Negative \(w\)
• Heat enters → Positive \(q\)
• Heat leaves → Negative \(q\)

Think of internal energy like the balance in a bank account. Heat entering and work done on the system are deposits, while heat leaving and work done by the system are withdrawals. The net balance gives \(\Delta E\).