Internal Energy, Heat, and Work Thermodynamics, Pressure & Volume, Chemistry Problems

Summary

This educational video script explores chemistry problems involving internal energy, heat, and work. It explains the first law of thermodynamics, using the formula ΔU = q + w, where ΔU is the change in internal energy, q is heat absorbed or released, and w is work done on or by the system. The script clarifies that q is positive for heat absorption and negative for release, while w is positive when work is done on the system and negative when work is done by the system. It provides examples to calculate internal energy changes in various scenarios, including heat absorption, work done, and gas expansion or compression against external pressure. The script also covers the conversion of work units from liters times atm to joules.

Takeaways

• 🔍 The video focuses on chemistry problems involving internal energy, heat, and work, emphasizing the application of the first law of thermodynamics.
• 🌡️ The change in internal energy (ΔU) is calculated using the formula ΔU = q + w, where q is the heat energy and w is the work done on or by the system.
• ♨️ Heat energy (q) is positive when absorbed by the system (endothermic process) and negative when released by the system (exothermic process).
• 🔨 Work (w) is positive when done on the system, increasing its internal energy, and negative when done by the system, decreasing its internal energy.
• 📈 The video explains the sign conventions for q and w in relation to the system and surroundings, crucial for correctly calculating ΔU.
• 🧪 Two examples are provided to illustrate the calculation of ΔU, one where the system absorbs heat and work is done on it, and another where the system releases heat and does work on the surroundings.
• 🌐 The video uses visual aids like diagrams to help understand the transfer of energy between the system and surroundings.
• 📉 The formula for work done by a gas during expansion or compression is w = -pΔV, where p is the constant external pressure, and ΔV is the change in volume.
• 🔄 The concept of energy storage in gases through pressure is discussed, highlighting that compressing a gas increases its internal energy, while expansion decreases it.
• 🔗 The video provides a conversion factor between liters-atm and joules to facilitate the calculation of work in terms of energy.
• 🔋 A final example calculates the internal energy change of a gas that absorbs heat and expands against a constant external pressure, demonstrating the application of the concepts discussed.

Q & A

What is the change in internal energy (ΔU) of a system when 300 joules of heat energy is absorbed and 400 joules of work is done on the system?

The change in internal energy (ΔU) can be calculated using the formula ΔU = q + w. Since 300 joules of heat is absorbed (q = +300 J) and 400 joules of work is done on the system (w = +400 J), the change in internal energy is ΔU = 300 J + 400 J = 700 J.

How does the sign of q (heat) affect the internal energy of a system?

In the context of the first law of thermodynamics, q is positive when heat is absorbed by the system (endothermic process), leading to an increase in internal energy. Conversely, q is negative when heat is released by the system (exothermic process), resulting in a decrease in internal energy.

What is the significance of a positive w (work) in thermodynamics?

A positive w indicates that work is done on the system, which means the system's internal energy increases. This typically occurs when the system is being compressed or when an external force is applied to it.

If a system releases 700 joules of heat and does 300 joules of work, what is the change in its internal energy?

The change in internal energy (ΔU) for this scenario would be ΔU = q + w = -700 J + (-300 J) = -1000 J. This means the system loses 1000 joules of energy, which is consistent with the first law of thermodynamics stating that energy is conserved and transferred from one form to another.

What does it mean when the surroundings gain 250 joules of heat energy in relation to the system?

If the surroundings gain 250 joules of heat energy, it implies that the system has released that amount of heat energy to the surroundings. Therefore, for the system, q would be -250 J (negative because heat is leaving the system).

How is work performed by the surroundings different from work done on the system?

When work is performed by the surroundings, it means that the surroundings are doing work, losing energy, and transferring that energy to the system, making w positive for the system. Conversely, when work is done on the system, it gains energy, and the surroundings lose energy.

What is the formula used to calculate the work done by a gas during expansion or compression?

The work done by a gas during expansion or compression is calculated using the formula w = -pΔV, where p is the constant external pressure, and ΔV is the change in volume (final volume - initial volume). The negative sign indicates that work done by the gas (expansion) is considered negative, while work done on the gas (compression) is positive.

How can you convert work done in liters-atm to joules?

To convert work done from liters-atm to joules, use the conversion factor where 1 liter-atm equals 101.3 joules. Multiply the work in liters-atm by this factor to get the work in joules.

What happens to the internal energy of a gas when it expands against a constant external pressure?

When a gas expands against a constant external pressure, it does work on the surroundings, which results in a decrease in the gas's internal energy. This is because the gas is converting its internal energy into work done on the surroundings.

If 500 joules of heat energy is absorbed by a gas and it expands from 30 liters to 70 liters against a constant pressure of 2.8 atm, how much is the change in internal energy?

The change in internal energy (ΔU) can be calculated by adding the heat absorbed (q = +500 J) to the work done by the gas (w = -pΔV). The work done by the gas during expansion is w = -2.8 atm * (70 L - 30 L) = -112 L*atm. Converting this to joules gives w = -112 * 101.3 J/L*atm = -11,345.6 J. Therefore, ΔU = 500 J + (-11,345.6 J) = -10,845.6 J.