Question

An ideal gas undergoes a constant temperature (isothermal) process in a closed system. The heat transfer and work are, respectively \((a) 0,-c_{v} \Delta T\) (b) \(c_{v} \Delta T, 0\) (c) \(c_{p} \Delta T, R \Delta T\) (d) \(R \ln \left(T_{2} / T_{1}\right), R \ln \left(T_{2} / T_{1}\right)\)

Short answer

a) (Q, W) = (0, -c_v ΔT) b) (Q, W) = (c_v ΔT, 0) c) (Q, W) = (c_p ΔT, R ΔT) d) (Q, W) = (R ln(T2 / T1), R ln(T2 / T1)) Answer: d) (Q, W) = (R ln(T2 / T1), R ln(T2 / T1))

Step-by-step solution

Apply First Law of Thermodynamics in an Isothermal Process

As the process is isothermal, the change in internal energy (\(\Delta U\)) is zero. Using the first law of thermodynamics for ideal gas in an isothermal process: \(Q = W\) Now, we will examine each option and see if it satisfies the condition \(Q = W\).

Examine Option A

Option (a) states \((Q, W) = (0, -c_v \Delta T)\), but the temperature is constant in an isothermal process. \(\Delta T\) should be zero. Thus, the given option contradicts the condition \(Q = W\).

Examine Option B

Option (b) states \((Q, W) = (c_v \Delta T, 0)\), but the temperature is constant in an isothermal process. \(\Delta T\) should be zero. Thus, the given option contradicts the condition \(Q = W\).

Examine Option C

Option (c) states \((Q, W) = (c_p \Delta T, R \Delta T)\), but the temperature is constant in an isothermal process. \(\Delta T\) should be zero. Thus, the given option contradicts the condition \(Q = W\).

Examine Option D

Option (d) states \((Q, W) = (R \ln \left(T_2 / T_1\right), R \ln \left(T_2 / T_1\right))\). The heat transfer and work done in this option are equal, satisfying the condition \(Q = W\). Since the process is isothermal, \(T_1=T_2\). Therefore, the answer is option (d) (\((Q, W) = (R \ln \left(T_2 / T_1\right), R \ln \left(T_2 / T_1\right))\)).

Key concepts

First Law of Thermodynamics

Understanding the first law of thermodynamics is essential for studying processes that involve energy changes in systems. Simplified, it states that energy can neither be created nor destroyed, only transferred or converted from one form to another. In a thermodynamic system, this translates to the principle that the change in internal energy of the system is equal to the heat added to the system minus the work done by the system on the surroundings.

Mathematically, it's expressed as: \( \[\begin{equation} \begin{aligned} \triangle U = Q - W \text{,} \text{ where }\Delta U \text{ is the change in internal energy, } Q \text{ is the heat added, and } W \text{ is the work done.} \text{ In the context of an isothermal process for an ideal gas, the temperature remains constant, indicating no change in internal energy (}\Delta U = 0 \text{), leading to the relationship } Q = W \text{.}\end{aligned} \end{equation}\] \)
When analyzing problems that feature these concepts, students should consider identifying the type of process (isothermal, adiabatic, etc.) first, which helps clarify the relationships between heat, work, and changes in internal energy.

Heat Transfer

Heat transfer, in thermodynamics, refers to the movement of energy from one place to another or from one system to another, due to temperature differences. There are three primary modes of heat transfer: conduction, convection, and radiation.

During an Isothermal Process

In an isothermal process, specifically for an ideal gas, the system's temperature remains constant. Hence, any heat transfer that occurs is used to do work on the surrounding environment. This relationship can be tricky because while the internal energy remains unchanged due to constant temperature, the heat transfer equals the work done by or on the gas. It emphasizes the concept that heat energy and work are two sides of the same coin in a thermodynamic process.

Internal Energy Change

The change in internal energy (\(\triangle U \)) of a thermodynamic system is a vital component of the first law. It's directly linked to the total energy contained within the system's molecules. For an ideal gas, the internal energy is only a function of the temperature.

Considering an Ideal Gas

If the temperature doesn't change, as in an isothermal process, there's effectively no change in kinetic energy of the gas molecules, and thus no change in internal energy (\(\triangle U = 0 \)). It's important to note that even when the energy of a system appears static (such as a constant temperature condition), energy transfer can still occur in the form of work, which reflects the versatility of the first law and the concept of energy conservation.

Ideal Gas Behavior

Ideal gases are hypothetical gases whose molecules occupy negligible space and have no interactions, and which strictly obey the ideal gas law: \(PV = nRT\), 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 absolute temperature. This behavior provides a good approximation for real gases at high temperature and low pressure.

Isothermal Expansion of an Ideal Gas

In an isothermal process, as the ideal gas expands, it does work on its surroundings. According to the ideal gas behavior, if the temperature is held constant during the expansion or compression of the gas, the product of pressure and volume remains constant. That's illustrated by the equation \(PV = \text{constant}\). However, to find the work done during such an isothermal process, one uses the equation \(W = nRT \text{ln}\left(\frac{V_2}{V_1}\right)\), which indicates the work done is path-dependent. In a nutshell, ideal gas behavior helps predict how energy transfer and work will take place in an isothermal process.