Problem 12 - University Physics with Modern Physics

Question

A gas in a cylinder is held at a constant pressure of \(2.30 \times 10^{5} \mathrm{Pa}\) and is cooled and compressed from 1.70 \(\mathrm{m}^{3}\) to \(1.20 \mathrm{m}^{3} .\) The internal energy of the gas decreases by \(1.40 \times 10^{5} \mathrm{J}\) (a) Find the work done by the gas. (b) Find the absolute value \(|Q|\) of the heat flow into or out of the gas, and state the direction of the heat flow. (c) Does it matter whether the gas is ideal? Why or why not?

Step-by-Step Solution

Answer

(a) \(-1.15 \times 10^{5} \text{ J}\); (b) \(2.55 \times 10^{5} \text{ J}\), heat flows out; (c) Gas type does not matter.

Step 1: Identify the Given Values

We are given that the pressure \( P \) of the gas is constant at \( 2.30 \times 10^{5} \, \text{Pa} \). The initial volume \( V_i \) is \( 1.70 \, \text{m}^{3} \) and the final volume \( V_f \) is \( 1.20 \, \text{m}^{3} \). The change in internal energy \( \Delta U \) is \(-1.40 \times 10^{5} \, \text{J}\).

Step 2: Calculate the Work Done by the Gas

The work done \( W \) by the gas is given by the formula: \[ W = P \cdot (V_f - V_i) \]Substitute the values:\[ W = 2.30 \times 10^{5} \, \text{Pa} \cdot (1.20 \, \text{m}^{3} - 1.70 \, \text{m}^{3}) \]\[ W = 2.30 \times 10^{5} \, \text{Pa} \cdot (-0.50 \, \text{m}^{3}) \]\[ W = -1.15 \times 10^{5} \, \text{J} \]This means the work done by the gas is \(-1.15 \times 10^{5} \, \text{J}\), and the negative sign indicates work is done on the gas.

Step 3: Apply the First Law of Thermodynamics

The first law of thermodynamics states that:\[ \Delta U = Q - W \]Solving for \( Q \), the heat flow:\[ Q = \Delta U + W \]Substitute the known values:\[ Q = -1.40 \times 10^{5} \, \text{J} + (-1.15 \times 10^{5} \, \text{J}) \]\[ Q = -2.55 \times 10^{5} \, \text{J} \]

Step 4: Determine the Direction of Heat Flow

The negative \( Q \) value indicates that heat flows out of the gas. Therefore, the absolute value of the heat flow is \(|Q| = 2.55 \times 10^{5} \, \text{J}\), and heat flows out of the gas.

Step 5: Consider If Gas Type Matters

The identity of the gas does not affect the solution because the process depends only on the initial and final states, and pressure is assumed constant. Whether the gas is ideal does not change the calculation of work or heat flow under these conditions.

Key Concepts

Work Done by Gas

In thermodynamics, when we talk about "work done by gas," we're referring to the energy transferred by the gas when it expands or is compressed against a pressure.
For a gas, the work done can be calculated using the formula:

• \( W = P \cdot (V_f - V_i) \)

Here, \( W \) is the work done, \( P \) is the constant pressure, and \( V_f - V_i \) is the change in volume.
In the given exercise, the gas is compressed, meaning the final volume \( V_f \) is less than the initial volume \( V_i \), which results in a negative work value \( (-1.15 \times 10^5 \, \text{J}) \).
This negative sign signifies that work is being done on the gas rather than by the gas. So, even though the gas itself isn't actively doing work (like expanding), outside forces are doing work on it to compress it.

Heat Flow

Heat flow (\( Q \)) is a fundamental concept in thermodynamics that refers to the transfer of thermal energy in and out of a system.
In this problem, we need to determine if the gas is gaining or losing heat using the first law of thermodynamics.

• The first law of thermodynamics can be expressed as: \( \Delta U = Q - W \)

This formula tells us that the change in internal energy (\( \Delta U \) ) of a system is equal to the heat added to the system, minus the work done by the system.
By rearranging this equation, we find:

• \( Q = \Delta U + W \)

Substituting in the given values:

• \( \Delta U = -1.40 \times 10^5 \, \text{J} \)
• \( W = -1.15 \times 10^5 \, \text{J} \)
• \( Q = -2.55 \times 10^5 \, \text{J} \)

The negative sign indicates that heat is leaving the system, which means heat is flowing out of the gas.

Internal Energy Change

Internal energy change (\( \Delta U \)) refers to the change in the total energy contained within the gas. This can happen through changes in temperature, phase, or volume.
In the context of the first law of thermodynamics, the change in internal energy is directly tied to both the work done and heat transfer.
The equation\( \Delta U = Q - W \) summarizes this relationship.
When the internal energy of the gas decreases, as in this exercise, it implies that either work is done on the gas, heat is lost, or both.
Here, with a decrease of \(-1.40 \times 10^5 \, \text{J}\), we know:

• The compression of the gas represents work done on it
• Negative heat flow (\(-2.55 \times 10^5 \, \text{J}\)) suggests heat was removed

Despite changes, this problem's context shows that the nature of the gas (ideal or real) doesn’t impact these calculations much, as these depend mainly on pressures and volumes involved.