The equation of state for one mole of a gas is \(P V=R T+B P\), where \(B\) is a constant, independent of temperature. The internal energy of fixed amount of gas is function of temperature only. If one mole of the above gas is isothermally expanded from \(12 \mathrm{~L}\) to \(22 \mathrm{~L}\) at a constant external pressure of 1 bar at \(400 \mathrm{~K}\), then the change in enthalpy of the gas is approximately \((\mathrm{B}=2 \mathrm{~L} / \mathrm{mol})\) (a) 0 (b) \(-3.32 \mathrm{~J}\) (c) \(-332 \mathrm{~J}\) (d) \(-166 \mathrm{~J}\)

Short Answer

Expert verified
-1000 J

Step by step solution

01

Understanding the provided information

We know the equation of state for our gas is given by the modified form of the ideal gas law: \(PV = RT + BP\). The problem states that the internal energy of the gas depends only on its temperature, and since the process is isothermal, the internal energy does not change. The process is furthermore done at constant external pressure.
02

Identifying the First Law of Thermodynamics

The First Law of Thermodynamics states that \(\Delta U = Q - W\), where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added to the system, and \(W\) is the work done by the system. Since the internal energy does not change during an isothermal process (\(\Delta U = 0\)), we have \(Q = W\).
03

Calculating the work done by the gas

For an isothermal process at constant external pressure, the work done by the gas, \(W\), is given by \(W = P_\text{ext} \cdot \Delta V\), where \(P_\text{ext}\) is the external pressure and \(\Delta V\) is the change in volume. We were given \(\Delta V = 22 \text{ L} - 12 \text{ L} = 10 \text{ L}\) and \(P_\text{ext} = 1 \text{ bar}\). Since 1 bar = \(10^5\) Pa and 1 L = \(10^{-3}\) m^3, we convert \(\Delta V\) into cubic meters to find work in Joules: \(W = 1 \times 10^5 \text{ Pa} \times 10 \times 10^{-3} \text{ m}^3 = 10^3 \text{ J}\).
04

Understanding Enthalpy change in terms of heat transfer

The enthalpy change \(\Delta H\) for an isothermal process at constant pressure equals the heat added to the system which is equal to the work done: \(\Delta H = Q\). Since there are no changes in internal energy for this isothermal process, \(Q = W\). Therefore, \(\Delta H = W\).
05

Correcting for additional pressure-volume work

However, the equation of state for the gas includes an additional term \(BP\), which contributes to the pressure-volume work. The additional work, \(W_{BP}\), is given by \(W_{BP} = \int_{V_1}^{V_2} BP dV = B \cdot P_{\text{ext}} \cdot \Delta V\). Substituting the given values, \(W_{BP} = 2 \text{ L/mol} \times 1 \text{ bar} \times 10 \text{ L} = 20 \text{ L bar}\). To convert L bar to Joules, we use 1 L bar = 100 J, hence \(W_{BP} = 20 \times 100 \text{ J} = 2000 \text{ J}\).
06

Calculating the change in enthalpy (\(\Delta H\))

Since we found that \(W = 1000 \text{ J}\) and \(W_{BP} = 2000 \text{ J}\), the total work done is the algebraic sum of these two, which is \(-1000 \text{ J}\). Therefore, the correction factor turns the work negative, and the change in enthalpy becomes \(\Delta H = -1000 \text{ J}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

First Law of Thermodynamics
Understanding the First Law of Thermodynamics is key to solving problems involving heat and work in thermodynamic processes. This fundamental law states that energy cannot be created or destroyed in an isolated system. More formally, it is expressed as \(\Delta U = Q - W\), where \(\Delta U\) is the change in internal energy of the system, \(Q\) is the heat added to the system, and \(W\) is the work done by the system.
In the context of our textbook problem, applying the First Law during an isothermal (constant temperature) expansion, where the internal energy does not change (\(\Delta U = 0\)), leads to the conclusion that the heat added to the gas is equal to the work done by it (\(Q = W\)). This provides the foundation to calculate heat and work for processes, which are further used to deduce other thermodynamic quantities such as enthalpy.
Isothermal Expansion
Isothermal expansion is a thermodynamic process that occurs at a constant temperature. This aspect is crucial to our gas expansion problem, where the temperature is maintained at 400 K. During isothermal processes, the internal energy of an ideal gas remains unchanged because it only depends on temperature. Hence, the change in internal energy (\(\Delta U\)) is zero.
Isothermal processes are characterized by a distinct relationship between pressure and volume, which follows from the ideal gas law. The work done in an isothermal expansion can generally be calculated using the formula \(W = nRT \ln(\frac{V_2}{V_1})\), where \(n\), \(R\), \(T\), \(V_1\), and \(V_2\) are the number of moles, the universal gas constant, the absolute temperature, the initial volume, and the final volume, respectively. However, due to our exercise involving an additional \(BP\) term in the equation of state, we must account for this additional pressure-volume work while calculating the total work done during expansion.
Pressure-Volume Work
Pressure-volume work, also known as \(PdV\) work, is the work done by or against a gas when it expands or contracts against a pressure. In thermodynamics, work can be thought of as the energy transfer that expands or compresses a gas.
In the problem, we are specifically looking at the work done by the gas during an isothermal expansion at constant external pressure. It is given by \(W = P_{\text{ext}} \cdot \Delta V\), where \(P_{\text{ext}}\) is the external pressure applied to the gas and \(\Delta V\) is the change in volume. Additionally, since our exercise includes the \(BP\) term in the gas's equation of state, the total work done on the system is not just given by \(P_{\text{ext}} \cdot \Delta V\), but also includes the work due to this extra term. Mathematically, this additional work can be represented as \(W_{BP} = \int_{V_1}^{V_2} BP dV = B \(P_{\text{ext}}\) \cdot \Delta V\), which signifies the additional energy considerations required for gases that deviate from ideal behavior.
Equation of State
An equation of state is a mathematical model that describes the state of matter under a given set of physical conditions. It provides a relationship between thermodynamic quantities such as pressure \(P\), volume \(V\), and temperature \(T\). The most renowned example is the Ideal Gas Law \(PV = nRT\), which describes the behavior of ideal gases.
In the given exercise, the gas is described by a modified equation of state \(PV = RT + BP\). This tells us that the gas does not behave ideally since the \(BP\) term modifies the ideal behavior captured by the simple \(PV\) relationship. The \(B\) factor is an indicator of the gas's deviation from ideality, independent of temperature. Variations in the equation of state from the Ideal Gas Law are vital for understanding real gas behavior, especially when predicting the work done during processes like expansion or compression, and they are crucial when calculating changes in enthalpy or other thermodynamic properties.

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