Question

A monatomic ideal gas expands isothermally from \(\left\\{p_{1}, V_{1}, T_{1}\right\\}\) to \(\left\\{p_{2}, V_{2}, T_{1}\right\\} .\) Then it undergoes an isochoric process, which takes it from \(\left\\{p_{2}, V_{2}, T_{1}\right\\}\) to \(\left\\{p_{1}, V_{2}, T_{2}\right\\}\) Finally the gas undergoes an isobaric compression, which takes it back to \(\left\\{p_{1}, V_{1}, T_{1}\right\\}\) a) Use the First Law of Thermodynamics to find \(Q\) for each of these processes. b) Write an expression for total \(Q\) in terms of \(p_{1}, p_{2}, V_{1},\) and \(V_{2}\).

Short answer

Based on the solution above, find the expression for the total heat transfer (Q) for the entire cyclic process in terms of given variables. Answer: \(Q_{Total} = -nRT_{1}\ln{\frac{V_{2}}{V_{1}}} - P_{1}(V_{1}-V_{2})\)

Step-by-step solution

Recall Ideal Gas Law and the First Law of Thermodynamics

To proceed, we must first recall the ideal gas law which is given by \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature. Then, we need to recall the First Law of Thermodynamics, which states that \(\Delta{U}= Q + W\), where \(\Delta{U}\) is the change in internal energy, Q is the heat transfer, and W is the work done on or by the gas.

Find Q for Isothermal Expansion

For the isothermal expansion, we have \(T_{1}=T_{1}\) (no temperature change) and \(P_{1}V_{1} = P_{2}V_{2}\) (from the ideal gas law, since the number of moles and temperature are constant). Also, the change in internal energy is \(\Delta{U} = \frac{3}{2} nR\Delta{T} = 0\), as there is no change in temperature. So, the First Law of Thermodynamics for this process becomes: \(Q_{1} = -W_{1}\), where \(W_{1}\) is the work done during the isothermal expansion. Work done during the isothermal process is given by \(W_{1} = nRT_{1}\ln{\frac{V_{2}}{V_{1}}}\). Hence, \(Q_{1} = -nRT_{1}\ln{\frac{V_{2}}{V_{1}}}\).

Find Q for Isochoric Process

For the isochoric process, we have \(V_{2}=V_{2}\) (no volume change). Also, as there is no volume change, no work is done during this step, \(W_{2}=0\). So the First Law of Thermodynamics for this process becomes: \(Q_{2} = \Delta{U_{2}}\), where \(\Delta{U_{2}} = \frac{3}{2} nR\Delta{T_{2}} = \frac{3}{2} nR(T_{2}-T_{1})\). Hence, \(Q_{2} = \frac{3}{2} nR(T_{2}-T_{1})\).

Find Q for Isobaric Compression

For the isobaric compression, we have \(P_{1}=P_{1}\) (no pressure change). Work done during the isobaric process is given by \(W_{3} = P_{1}(V_{1}-V_{2})\). The change in internal energy is \(\Delta{U_{3}} = \frac{3}{2} nR\Delta{T_{3}} = \frac{3}{2} nR(T_{1}-T_{2})\) (since we are going back to the initial state). Hence, by the First Law of Thermodynamics, \(Q_{3} = \Delta{U_{3}} - W_{3} = \frac{3}{2} nR(T_{1}-T_{2}) - P_{1}(V_{1}-V_{2})\).

Find Total Q

Now we have expressions for the heat transfer during each process (\(Q_{1}, Q_{2}\), and \(Q_{3}\)). To find the total heat transfer for the entire cycle, we just need to add them: \(Q_{Total} = Q_{1} + Q_{2} + Q_{3} = -nRT_{1}\ln{\frac{V_{2}}{V_{1}}} + \frac{3}{2} nR(T_{2}-T_{1}) + \frac{3}{2} nR(T_{1}-T_{2}) - P_{1}(V_{1}-V_{2})\). After some simplification, we get: \(Q_{Total} = -nRT_{1}\ln{\frac{V_{2}}{V_{1}}} - P_{1}(V_{1}-V_{2})\).

Key concepts

Ideal Gas Law

Understanding the behavior of gases is essential in thermodynamics, and the ideal gas law is a fundamental equation that relates the pressure (P), volume (V), and temperature (T) of a hypothetical ideal gas. It is expressed as:
\( PV = nRT \),
where n represents the number of moles of the gas and R is the universal gas constant. This law assumes that the gas particles do not exert forces on each other and occupy no volume, which isn't entirely accurate but is a good approximation for many gases under normal conditions.
In our exercise, this law is used to analyze changes in the state of the gas during the different processes it undergoes.

Isothermal Expansion

The term isothermal refers to a process that occurs at a constant temperature. During an isothermal expansion, a gas expands, and its volume increases while its temperature remains the same. According to the ideal gas law, if temperature and quantity of gas remain constant, the product of pressure and volume for the gas must also remain constant. This provides a basis for calculating work done by the gas during this process, which is important when applying the First Law of Thermodynamics to determine the heat transfer, Q.
For the isothermal process in our exercise, we see that the work done and heat transfer are equal in magnitude but opposite in sign.

Isochoric Process

In contrast to expansion or compression, an isochoric process involves no change in volume. The term 'isochoric' stems from Greek, where 'iso-' means same and 'choric' relates to space. Since the volume doesn't change, there is also no work done by the gas (work is often a product of pressure and volume change). Therefore, any heat added or removed from the gas goes into changing its internal energy. This can be observed in the exercise where an increase in heat results in a rise in temperature without altering the volume.

Isobaric Compression

Compression where the pressure remains constant is known as an isobaric compression. The term 'isobaric' is derived from 'iso-' meaning same and 'baric' pertaining to pressure. Work is involved in this process since there is a volume change at a constant pressure. The First Law of Thermodynamics helps us quantify the heat transfer involved in returning the gas to its original volume and temperature. In the exercise, we used this concept to calculate the work done on the gas and the associated heat transfer, Q, during the compression.

Heat Transfer in Thermodynamics

In the realm of thermodynamics, heat transfer signifies the exchange of thermal energy between physical systems. The concept is based on the Second Law of Thermodynamics, which infers that heat flows naturally from hot to cold bodies. There are three primary modes of heat transfer: conduction, convection, and radiation. However, in our exercise example, we focus on the heat quantity (Q) in the context of the First Law of Thermodynamics. This law correlates heat transfer to changes in a gas's internal energy and the work done by or on the gas, thereby laying the foundation for solving the given problem.

Work Done by Gas

The work done by a gas is a significant concept because it links the energy change with mechanical actions. Work is defined as force exerted over a distance. In thermodynamics, when a gas expands or is compressed, it either does work on its surroundings or has work done on it, respectively. The work can be calculated by interfacing pressure with volume changes of the gas.
In the given exercise, the work done is a vital part of calculating Q in various processes. Notably, an isothermal expansion leads to work done by the gas, while an isobaric compression refers to work done on the gas. These distinctions are crucial when applying the First Law of Thermodynamics to determine heat transfer.