Ch.19 - Chemical Thermodynamics

Brown - Chemistry: The Central Science 14th Edition

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Brown 14th Edition

Ch.19 - Chemical Thermodynamics

Problema 29b

Capitolo 19, Problema 29b

For the isothermal expansion of a gas into a vacuum, ΔE = 0, q = 0, and w = 0. (b) Explain why no work is done by the system during this process.

Guida verificata passo dopo passo

insert step 1> Understand the concept of isothermal expansion into a vacuum, which is also known as free expansion.

insert step 2> Recognize that in a vacuum, there is no external pressure opposing the expansion of the gas.

insert step 3> Recall that work done by a gas during expansion is given by the formula: w = -P_{ext} \(\Delta\) V, where P_{ext} is the external pressure.

insert step 4> Since the gas is expanding into a vacuum, P_{ext} = 0, meaning there is no opposing force.

insert step 5> Conclude that with P_{ext} = 0, the work done, w, is also 0, as w = -0 \(\times\) \(\Delta\) V = 0.

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Isothermal Process

An isothermal process occurs at a constant temperature, meaning that the internal energy of an ideal gas remains unchanged during the expansion or compression. In this scenario, since the temperature is constant, any heat added to the system is used to do work, but in the case of expansion into a vacuum, there is no heat exchange with the surroundings.

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Spontaneity of Processes

Work in Thermodynamics

In thermodynamics, work is defined as the energy transfer that occurs when a force is applied over a distance. For a gas expanding against an external pressure, work is done by the system. However, in an expansion into a vacuum, there is no opposing pressure, meaning the gas does not push against anything, resulting in zero work done.

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First Law of Thermodynamics

The First Law of Thermodynamics states that energy cannot be created or destroyed, only transformed. In the case of the isothermal expansion into a vacuum, the change in internal energy (ΔE) is zero, indicating that the energy remains constant. Since no work is done and no heat is exchanged, the system's energy balance confirms that ΔE = q + w = 0.

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First Law of Thermodynamics

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(a) What sign for ΔS do you expect when the pressure on 0.600 mol of an ideal gas at 350 K is increased isothermally from an initial pressure of 0.750 atm? (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?

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(a) What is the difference between a state and a microstate of a system?

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(c) In a particular spontaneous process, the number of microstates available to the system decreases. What can you conclude about the sign of ΔS_surr?

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(b) As a system goes from state A to state B, its entropy decreases. What can you say about the number of microstates corresponding to each state?

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