Calculating Internal Energy Change in Gas Compression

This problem involves a stationary mass of gas undergoing a thermodynamic process. We are asked to find the change in internal energy of the gas given the initial and final states, the heat transfer, and the type of process.

Understanding the First Law of Thermodynamics

The First Law of Thermodynamics is a fundamental principle that relates the change in internal energy ($\Delta U$) of a system to the heat added to the system ($Q$) and the work done by the system ($W$). 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. If heat is transferred from the system, \(Q\) is negative.
• \(W\) is the work done by the system. If work is done on the system (like in compression), \(W\) is negative.

Analyzing the Given Information

We are given the following details about the gas and the process:

• Initial volume, \(V_1 = 0.3 \, \text{m}^3\)
• Initial pressure, \(P_1 = 0.105 \, \text{MPa}\)
• Final volume, \(V_2 = 0.15 \, \text{m}^3\)
• Final pressure, \(P_2 = 0.105 \, \text{MPa}\)
• The pressure remains constant during the process, meaning it is an isobaric process. \(P = P_1 = P_2 = 0.105 \, \text{MPa}\)
• Heat transferred from the gas, \(Q_{out} = 37.6 \, \text{kJ}\). Since heat is leaving the system, the heat added to the system is \(Q = -37.6 \, \text{kJ}\).

We need to calculate \(\Delta U\).

Calculating the Work Done During Constant Pressure Compression

For a process where the pressure remains constant (isobaric process), the work done by the system is given by the formula:

\(W = P(V_2 - V_1)\)

First, let's convert the pressure from MPa to Pa:

\(P = 0.105 \, \text{MPa} = 0.105 \times 10^6 \, \text{Pa}\)

Now, substitute the values for pressure, initial volume, and final volume into the work equation:

\(W = (0.105 \times 10^6 \, \text{Pa}) \times (0.15 \, \text{m}^3 - 0.3 \, \text{m}^3)\)

\(W = (0.105 \times 10^6) \times (-0.15) \, \text{J}\)

\(W = -0.01575 \times 10^6 \, \text{J}\)

\(W = -15750 \, \text{J}\)

To convert Joules to kilojoules, divide by 1000:

\(W = -15.75 \, \text{kJ}\)

The work done by the gas is negative, which is expected for a compression process (work is done on the gas).

Calculating the Change in Internal Energy

Now that we have the heat transfer \(Q\) and the work done \(W\), we can use the First Law of Thermodynamics to find the change in internal energy \(\Delta U\).

\(\Delta U = Q - W\)

Substitute the values \(Q = -37.6 \, \text{kJ}\) and \(W = -15.75 \, \text{kJ}\):

\(\Delta U = -37.6 \, \text{kJ} - (-15.75 \, \text{kJ})\)

\(\Delta U = -37.6 \, \text{kJ} + 15.75 \, \text{kJ}\)

\(\Delta U = -21.85 \, \text{kJ}\)

Conclusion on Internal Energy Change

The change in the internal energy of the gas during this constant pressure compression process is -21.85 kJ. A negative change in internal energy means that the internal energy of the gas has decreased.