Problem 31 - University Physics with Modern Physics

Question

An experimenter adds \(970 \mathrm{~J}\) of heat to \(1.75 \mathrm{~mol}\) of an ideal gas to heat it from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C}\) at constant pressure. The gas does \(+223 \mathrm{~J}\) of work during the expansion. (a) Calculate the change in internal energy of the gas. (b) Calculate \(\gamma\) for the gas.

Step-by-Step Solution

Answer

(a) The change in internal energy is 747 J. (b) The value of \( \gamma \) is approximately 1.30.

Step 1: Understand the First Law of Thermodynamics

For any thermodynamic process, the change in internal energy (\( \Delta U \)) of a system is related to the heat added to the system (\( Q \)) and the work done by the system (\( W \)) by the formula:\( \Delta U = Q - W \).

Step 2: Calculate the Change in Internal Energy

We are given \( Q = 970 \mathrm{~J} \) and \( W = 223 \mathrm{~J} \). Substitute these values into the formula:\[\Delta U = 970 \mathrm{~J} - 223 \mathrm{~J} = 747 \mathrm{~J}\]The change in internal energy of the gas is \( 747 \mathrm{~J} \).

Step 3: Understand Heat Capacity and Gamma (γ) Relationship

The heat capacity ratio (\( \gamma \)) is defined as \( \gamma = \frac{C_P}{C_V} \), where \( C_P \) is the heat capacity at constant pressure, and \( C_V \) is the heat capacity at constant volume. We can use the molar heat capacities for ideal gases to calculate \( \gamma \). The relationship for ideal gases at constant pressure is given as \( \Delta Q = n C_P \Delta T \), where \( n \) is the number of moles and \( \Delta T \) is the temperature change.

Step 4: Calculate Molar Heat Capacity at Constant Pressure (CP)

Given \( n = 1.75 \mathrm{~mol} \) and \( \Delta T = (25.0^{\circ} \mathrm{C} - 10.0^{\circ} \mathrm{C}) = 15.0 \mathrm{~K} \), substitute into \( \Delta Q = n C_P \Delta T \) to solve for \( C_P \).\[970 \mathrm{~J} = 1.75 \mathrm{~mol} \times C_P \times 15.0 \mathrm{~K}\]\[C_P = \frac{970}{1.75 \times 15.0} = \frac{970}{26.25} \approx 36.95 \frac{\mathrm{J}}{\mathrm{mol \cdot K}}\]

Step 5: Calculate Molar Heat Capacity at Constant Volume (CV)

From \( \Delta U = n C_V \Delta T \), given that \( \Delta U = 747 \mathrm{~J} \), we solve for \( C_V \).\[747 \mathrm{~J} = 1.75 \mathrm{~mol} \times C_V \times 15.0 \mathrm{~K}\]\[C_V = \frac{747}{1.75 \times 15.0} = \frac{747}{26.25} \approx 28.46 \frac{\mathrm{J}}{\mathrm{mol \cdot K}}\]

Step 6: Calculate Gamma (γ)

Finally, substitute the calculated values into the formula \( \gamma = \frac{C_P}{C_V} \).\[\gamma = \frac{36.95}{28.46} \approx 1.30\]The value of \( \gamma \) for the gas is approximately \( 1.30 \).

Key Concepts

Internal Energy

Internal energy is a fundamental concept in thermodynamics, representing the total energy contained within a system. This energy comes from particles moving and interacting inside the substance. When you add heat to a system or it performs work, this energy can change.
The First Law of Thermodynamics describes how energy changes occur within any system. It is expressed as:

• \( \Delta U = Q - W \)

Here, \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system. In essence, this law is about conservation of energy: the energy that enters a system must equal the energy leaving it. Understanding how energy distribution between heat and work shifts the internal energy is crucial to solving related problems in thermodynamics.

Heat Capacity

Heat capacity defines how much heat a substance can store per degree of temperature increase. It's an important property in analyzing thermal processes of materials. Heat capacity is expressed in two common ways:

• Molar Heat Capacity (at constant pressure or volume): Amount of heat required to raise the temperature of one mole of a substance by one degree Kelvin.

For gases, we often deal with heat capacities at constant pressure (\( C_P \)) and constant volume (\( C_V \)). The relationship in our context is also used to determine \( \gamma \), the heat capacity ratio:

• \( \gamma = \frac{C_P}{C_V} \)

In an ideal gas, these values help predict how gases will respond to heat under different conditions, like whether the gas is expanding or being compressed while retaining its constant volume.

Ideal Gas

An ideal gas is a simplified model used in physics and chemistry to better understand the behavior of gases. It assumes that:

• The molecules do not attract or repel each other.
• The volume of the individual gas molecules is negligible.

Even though no real gases perfectly fit these assumptions, the ideal gas law is a useful approximation for calculating relationships involving temperature, pressure, and volume in real gases under many conditions. This law is expressed as:

• \( PV = nRT \)

Where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is the universal gas constant, and \( T \) is temperature. For an ideal gas, this equation allows you to link the properties of a gas through simple mathematical relationships, making calculations like those in our problem straightforward.

Work Done by Gas

When a gas expands or compresses, it does work on its surroundings. The work done by a gas is a form of energy transfer, important in understanding thermodynamic processes. It's calculated based on the change in volume and pressure of the gas.
In a constant pressure process (like our problem), work done by the gas can be given by:

• \( W = P \Delta V \)

Where \( W \) is the work done by the gas, and \( \Delta V \) is the change in volume. However, we often use different forms of this equation depending on the known variables, as in our example where work was already provided.
Knowing the work done, along with heat inputs or outputs, allows us to calculate changes in internal energy and analyze energy conversions within the system. Understanding this concept is essential for predicting and controlling the behavior of gases in engines, refrigerators, and other applications.