Problem 32 - Chemistry and Chemical Reactivity

Question

Nitrogen gas is confined in a cylinder with a movable piston under a constant pressure of \(9.95 \times 10^{4}\) Pa. When \(695 \mathrm{J}\) of energy in the form of heat is transferred from the gas to the surroundings, its volume decreases by 1.88 L. What is the change in internal energy of the gas?

Step-by-Step Solution

Answer

The change in internal energy of the gas is -508.14 J.

Step 1: Identify Known Variables

The problem provides the following known variables: - Pressure \( P = 9.95 \times 10^4 \text{ Pa} \)- Heat transferred \( Q = -695 \text{ J} \) (negative because energy is transferred from the gas to the surroundings)- Volume change \( \Delta V = -1.88 \text{ L} = -1.88 \times 10^{-3} \text{ m}^3 \) (converted to cubic meters)

Step 2: Use the First Law of Thermodynamics

The first law of thermodynamics states that\[ \Delta U = Q - W \]where \( \Delta U \) is the change in internal energy, \( Q \) is the heat exchanged, and \( W \) is the work done by/on the system.

Step 3: Calculate the Work Done on the System

Work done by or on a piston-system at constant pressure is given by\[ W = P \Delta V \]Substitute the known values:\[ W = 9.95 \times 10^4 \text{ Pa} \times (-1.88 \times 10^{-3} \text{ m}^3) \]\[ W = -186.86 \text{ J} \]

Step 4: Calculate the Change in Internal Energy

Substitute the values for \( Q \) and \( W \) into the equation from Step 2:\[ \Delta U = Q - W \]\[ \Delta U = (-695 \text{ J}) - (-186.86 \text{ J}) \]\[ \Delta U = -695 \text{ J} + 186.86 \text{ J} \]\[ \Delta U = -508.14 \text{ J} \]

Step 5: Conclusion: Result Interpretation

The change in internal energy \( \Delta U \) is \(-508.14 \text{ J} \). This means the internal energy of the gas has decreased by \( 508.14 \text{ J} \), indicating a loss of energy.

Key Concepts

Change in Internal Energy

The change in internal energy, denoted as \( \Delta U \), represents the total change in energy within a system due to heat transfer and work done by or on the system. In the context of the first law of thermodynamics, this expression is crucial because it combines the effects of heat \( Q \) and work \( W \) to provide an understanding of how energy dynamics within the system alter.
For a gas such as nitrogen in a piston, if \( \Delta U \) is negative, it means the internal energy has decreased, indicating that the system has lost energy. This is a direct outcome of the sum of heat transferred out and work done by the gas. Conversely, a positive \( \Delta U \) would suggest the system has gained energy. In this example, \( \Delta U \) is \(-508.14 \text{ J}\), showing the gas has indeed lost energy.

Work Done by/on the System

Work done by or on a system involves energy being transferred due to volume changes under pressure. For systems like the gas within a cylinder with a piston, work \( W \) is calculated using the formula \( W = P \Delta V \), where \( P \) is pressure and \( \Delta V \) is the change in volume.
In this problem, the negative volume change indicates the gas is doing work on the surroundings by pushing the piston inwards. The calculated work is \( -186.86 \text{ J} \), signifying the outgoing work done by the gas.
This concept helps in understanding that when a gas expands, it does positive work (energy leaves the system) and when compressed, it does negative work (energy enters the system), depending on the direction of volume change.

Heat Transfer

Heat transfer \( Q \) refers to the process of energy transfer due to temperature differences between the system and its surroundings. According to the given problem, \( 695 \text{ J} \) of heat is transferred from the gas to the surroundings, causing a loss of energy in the system.
This negative heat exchange (represented as \( -695 \text{ J} \)) hints that energy is moving out of the system, impacting its internal energy. When a system loses heat, it might also result in work being done, which together influences the overall energy balance.
Understanding heat transfer is essential as it provides insight into how energy enters or leaves a system purely due to thermal dynamics, affecting both temperature and the work capacity of the gas.