Question

A 1.25 mole sample of an ideal gas is expanded from \(320 . \mathrm{K}\) and an initial pressure of 3.10 bar to a final pressure of 1.00 bar, and \(C_{P, m}=5 R / 2 .\) Calculate \(w\) for the following two cases: a. The expansion is isothermal and reversible. b. The expansion is adiabatic and reversible. Without resorting to equations, explain why the result to part (b) is greater than or less than the result to part (a).

Short answer

In both cases, the ideal gas expands and does work on its surroundings. The work done in case a (isothermal expansion) is approximately \(-204.3 \mathrm{L\cdot bar}\), while the work done in case b (adiabatic expansion) is approximately \(-299.8 \mathrm{L\cdot bar}\). The work done in case b is greater in magnitude than the work done in case a because, in an adiabatic process, the gas uses its internal energy to do work, leading to a decrease in temperature. In contrast, in an isothermal process, the gas temperature remains constant throughout the expansion.

Step-by-step solution

Recall the formula for work in an isothermal process

In an isothermal and reversible expansion, the work done by the gas can be calculated using the formula: \[w = -nRT \ln\left(\frac{P_2}{P_1}\right)\] where \(n\) is the number of moles, \(R\) is the gas constant, \(T\) is the constant temperature, \(P_1\) is the initial pressure, and \(P_2\) is the final pressure.

Plug in the given values

We're given the following values: \(n = 1.25\) moles, \(T = 320 \mathrm{K}\), \(P_1 = 3.10 \mathrm{bar}\), and \(P_2 = 1.00 \mathrm{bar}\). The gas constant \(R\) has a value of \(8.314 \mathrm{J/mol \cdot K}\), but since the pressures are given in bar, we need to convert \(R\) to L·bar/mol·K using the relation: 1 L·bar = 100 J. So we get \(R = 0.08314 \mathrm{L \cdot bar/mol \cdot K}\). Now plug in the values and calculate the work: \[w = -(1.25 \mathrm{mol})(0.08314 \mathrm{L \cdot bar/mol \cdot K})(320 \mathrm{K}) \ln\left(\frac{1.00 \mathrm{bar}}{3.10 \mathrm{bar}}\right)\]

Calculate the work

Calculating the work done during the isothermal expansion, we get: \[w \approx -204.3 \mathrm{L\cdot bar}\] where the negative sign means the gas does work on the surroundings. Case b. Adiabatic and reversible expansion:

Recall the relation between the initial and final pressures and temperatures

In an adiabatic and reversible expansion, the following relation holds: \[\frac{P_1^{\gamma - 1}}{P_2^{\gamma - 1}} = \frac{T_2}{T_1}\] where \(\gamma = \frac{C_P}{C_V}\) is the adiabatic index.

Calculate the final temperature

First, we need to find the adiabatic index \(\gamma\). We're given that \(C_{P,m}=\frac{5R}{2}\), so \(C_{V,m} = C_{P,m} - R = \frac{3R}{2}\). Therefore, \(\gamma = \frac{C_{P,m}}{C_{V,m}}=\frac{5}{3}\) . Now, use the relation between pressures and temperatures to find the final temperature. Since the initial temperature \(T_1\) is known, we can find \(T_2\): \[T_2 = \frac{P_1^{\gamma - 1}}{P_2^{\gamma - 1}} \cdot T_1 = \frac{(3.10 \mathrm{bar})^{2/3}}{(1.00 \mathrm{bar})^{2/3}} \cdot 320 \mathrm{K}\]

Calculate the work

Now, we can use the formula for work in an adiabatic process: \[w = \frac{nR(T_1 - T_2)}{\gamma - 1}\] Plug in the values and calculate the work: \[w = \frac{1.25 \mathrm{mol} \cdot 0.08314 \mathrm{L \cdot bar/mol \cdot K} ( 320 \mathrm{K} - T_2 )}{2/3}\] After obtaining the value of \(T_2\) and calculating the work done during the adiabatic expansion, we find: \[w \approx -299.8 \mathrm{L\cdot bar}\] Comparison: The work done in case b (adiabatic expansion) is greater in magnitude than the work done in case a (isothermal expansion). This happens because, in an adiabatic process, no heat is gained (nor lost) from the surroundings. Therefore, for the gas to do work on the surroundings, it has to use its internal energy which results in a decrease in temperature. This means that in the adiabatic process, the gas is working against both the external pressure and its own decreasing temperature, requiring more work. On the other hand, in an isothermal process, the gas will do work against the external pressure, but its temperature will remain constant throughout the process.

Key concepts

Isothermal Process

An isothermal process occurs when a gas expands or compresses at a constant temperature. During this process, the temperature of the system does not change, illustrating the term 'isothermal', which literally means 'same temperature'. In the context of an ideal gas, this implies that the gas has to interact with a thermal reservoir (a body that can supply or absorb heat without changing its temperature) as it does work or work is done on it.

For a reversible isothermal expansion of an ideal gas, work is calculated using the formula: \[w = -nRT \ln\left(\frac{P_2}{P_1}\right)\]Where:

• n is the number of moles,
• R is the ideal gas constant,
• T is the constant temperature, and
• P_1 and P_2 are the initial and final pressures, respectively.

The negative sign indicates work is done by the gas on its surroundings. It's important to note that in an isothermal process, the internal energy of the gas remains constant due to the constant temperature. As a result, any work done by the gas is balanced by an equal amount of heat absorbed from the surroundings.

Reversible Expansion

A reversible expansion is a theoretical concept where changes happen so slowly and infinitesimally that the system is always in equilibrium with its surroundings. In other words, every intermediate state through which the system passes can be considered an equilibrium state. This concept is crucial in thermodynamics as it defines an idealized process against which real, irreversible processes can be compared.

In reversible expansion, not only can the process go forward in such a way that at each stage, the system is in perfect equilibrium, but it can also be reversed by infinitesimally small changes in control parameters like pressure and volume. This means that both the system and the surroundings can be returned to their original states without any net changes in the world occurring. It's this precise reversibility and adherence to equilibrium that allows for the maximum amount of work to be extracted during an expansion or minimized during a compression.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation that relates the pressure, volume, temperature, and amount (in moles) of an ideal gas. This law is a combination of Boyle's Law, Charles's Law, and Avogadro's Law and is expressed in the formula: \[PV = nRT\]Where P is the pressure of the gas, V is the volume, n is the number of moles of the gas, R is the universal gas constant, and T is the absolute temperature in Kelvin.

The Ideal Gas Law assumes that the gas molecules themselves occupy no volume and that there are no interactive forces between the molecules. While no real gas perfectly fits these criteria, under normal conditions, many gases behave very similarly to an ideal gas. Therefore, the Ideal Gas Law becomes a powerful tool for understanding and calculating the behavior of gases in a variety of chemical and physical processes.

Work Done by Gas

The work done by a gas during expansion or compression is a way of describing the energy transfer from the gas to its surroundings, or vice-versa. When a gas expands, it exerts a force over a distance, effectively doing work on the objects around it. Conversely, when a gas is compressed, the surroundings do work on the gas.

The calculation for work done by a gas depends on the specific process. For an isothermal, reversible process, the formula for work is: \[w = -nRT \ln\left(\frac{V_f}{V_i}\right)\ = -nRT \ln\left(\frac{P_i}{P_f}\right)\]For an adiabatic, reversible process—when no heat exchange takes place—the work is given by the differences in temperature and the number of moles of the gas as: \[w = \frac{nR(T_1 - T_2)}{\gamma - 1}\]Here, \[\gamma\] is the ratio of the heat capacities, which is constant for a given ideal gas. Understanding how work is done by a gas is critical in the study of thermodynamics as it relates to energy efficiency, engine cycles, refrigeration, and atmospheric science.