Problem 37 - University Physics with Modern Physics

Question

One mole of ideal gas is slowly compressed to one-third of its original volume. In this compression, the work done on the gas has magnitude 600 \(\mathrm{J}\) . For the gas, \(C_{p}=7 R / 2\) . (a) If the process is isothermal, what is the heat flow \(Q\) for the gas? Does heat flow into or out of the gas? (b) If the process is isobaric, what is the change in internal energy of the gas? Does the internal energy increase or decrease?

Step-by-Step Solution

Answer

(a) Heat flow is -600 J (out of the gas). (b) Internal energy change is +1500 J (increases).

Step 1: Analyze the Isothermal Process

In an isothermal process for an ideal gas, the temperature remains constant. From the First Law of Thermodynamics, the change in internal energy \(\Delta U = 0\) for an isothermal process. The work \(W\) done on the gas can thus be related to the heat flow \(Q\) by the equation: \[ Q = -W \]Given that the work done on the gas is 600 \(\mathrm{J}\) when compressed, the heat flow into the gas is \(-600 \; \mathrm{J}\). This means that heat flows out of the gas.

Step 2: Analyze the Isobaric Process

In an isobaric process, the pressure remains constant. For one mole of ideal gas, the specific heat \(C_p = \frac{7R}{2}\). The change in internal energy \(\Delta U\) for the process is given by:\[ \Delta U = nC_v\Delta T \]where \(C_v = C_p - R = \frac{5R}{2}\). Since the work done is given by the equation:\[ W = -P \Delta V = nR \Delta T \]Solve for \(\Delta T\): \[ W = nR \Delta T \rightarrow \Delta T = \frac{W}{nR} = \frac{600}{R} \]Substitute back to find \(\Delta U\):\[ \Delta U = n \cdot \frac{5R}{2} \cdot \frac{600}{R} = 1500 \; \mathrm{J} \]The internal energy increases.

Key Concepts

Ideal Gas

The ideal gas concept is fundamental in thermodynamics. An ideal gas is a hypothetical gas that follows the ideal gas law perfectly. This law relates the pressure (P), volume (V), temperature (T), and number of moles (n) of a gas through the equation:

• \( PV = nRT \)

Here, \( R \) is the universal gas constant. An ideal gas is assumed to have molecules that occupy no volume and experience no intermolecular forces. This simplification allows for easier calculations and predictions of behavior under various conditions.

• Real gases approximate ideal gases at high temperatures and low pressures.
• The concept helps in understanding various thermodynamic processes like isothermal and isobaric processes.

Isothermal Process

An isothermal process occurs when a gas undergoes changes at a constant temperature. For an ideal gas undergoing such a process, there is no change in internal energy (\( \Delta U = 0 \)). The relationship between work \( W \) and heat flow \( Q \) can be expressed using the First Law of Thermodynamics:

• \( \Delta U = Q - W \)

Since \( \Delta U = 0 \), it simplifies to:

• \( Q = W \)

In the context of a compression that increases the pressure, if work is done on the gas, the heat will typically flow out to maintain a constant temperature. This is seen in the given exercise, where for the isothermal compression, heat flow \( Q = -600 \) J, meaning heat exits the gas to compensate for the work done against it.

Isobaric Process

During an isobaric process, a gas changes its state while maintaining constant pressure. This differs from isothermal changes, as temperature can vary. The challenge here is to determine the change in internal energy \( \Delta U \) as the state of the gas alters. Using:

• \( \Delta U = nC_v\Delta T \)

And knowing \( C_p \) (specific heat at constant pressure), we calculate \( C_v \) (specific heat at constant volume) as:

• \( C_v = C_p - R = \frac{5R}{2} \)

Given the work value and pressure constancy, we find the temperature change \( \Delta T \), allowing us to compute internal energy change:

• \( \Delta U = 1500 \) J

In the exercise, internal energy increases due to the energy input to maintain pressure, representing work done on the gas.

Internal Energy

Internal energy is a state property of a system, indicating total energy contained within it (kinetic + potential). For an ideal gas, internal energy is only a function of temperature:

• \( U = nC_v T \)

Changes in internal energy (\( \Delta U \)) are crucial in thermodynamic cycles, especially in assessing systems' energy exchanges:

• In isothermal processes, internal energy change is zero, \( \Delta U = 0 \). Temperatures don’t change.
• In isobaric processes, \( \Delta U \) hinges on temperature variations induced by work or heat flow.

Tracking \( \Delta U \) is vital for energy conservation and system efficiency analyses.

Heat Flow

Heat flow (\( Q \)) is a measure of thermal energy transfer into or out of a system. Together with work, it forms a cornerstone of the First Law of Thermodynamics:

• \( \Delta U = Q - W \)

Describing energy conservation, heat flow impacts crucial processes:

• In isothermal actions, \( Q = W \), indicating direct replacement of energy by heat to balance work done.
• Isobarically, \( Q \) adjusts alongside work and temperature to hold pressure constant.

By observing heat flow direction and magnitude, one discerns whether a system absorbs or dispels energy to sustain equilibrium and analyze thermodynamic processes.