How much work is done by the steam when 1.00 mol of water...

How much work is done by the steam when $1.00 \mathrm{~mol}$ of water at $100^{\circ} \mathrm{C}$ boils and becomes $1.00 \mathrm{~mol}$ of steam at $100^{\circ} \mathrm{C}$ and at $1.00$ atm pressure? Assuming the steam to be an ideal gas, determine the change in internal energy of the steam as it vaporizes.

Key Concepts

Ideal Gas Law

This is the relationship among the pressure, volume, temperature, and number of moles of a gas, expressed as PV = nRT. It is used to determine how the volume changes when the gas is heated or expanded at a fixed pressure, which is essential for calculating the work done during an expansion process.

Work in Thermodynamic Processes

Work in thermodynamics, often expressed as W = P?V for processes at constant pressure, represents the energy transferred by the system as it expands or contracts. In phase change problems, such as boiling, the work done by the gas can be determined by the pressure and the change in volume when the liquid changes to vapor.

First Law of Thermodynamics

This fundamental principle states that the change in internal energy of a system, ?U, is equal to the heat added to the system, Q, minus the work done by the system, W (?U = Q - W). It is used to analyze energy changes during processes like vaporization where both heat and work contributions must be accounted for.

Internal Energy in Ideal Gases

For an ideal gas, the internal energy depends solely on the temperature of the gas and the degrees of freedom of its molecules. In an isothermal process, where the temperature remains constant, the internal energy does not change even if work is done by or on the gas. This concept is crucial when analyzing the final state of steam treated as an ideal gas during vaporization.

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