ISC Class 11 Chemistry – First Law of Thermodynamics Made Simple

Ever wondered why a hot cup of tea cools down on its own? That everyday mystery is actually a perfect showcase of the first law of thermodynamics.

💡 In Simple Words: Energy can’t be created or destroyed, it only moves around or changes form. So the total energy of the universe stays the same, even if the coffee gets cooler and the room gets a tiny bit warmer.

What is the First Law of Thermodynamics?

The first law is just a fancy way of saying energy conservation. In chemistry we write it as:

ΔU = q + w

Here ΔU (delta U) is the change in internal energy of the system – the tiny particles inside whatever you’re studying. q stands for heat added to the system, and w is work done on the system. If heat leaves or work is done by the system, those quantities become negative.

Key Terms Explained

• System: The part you’re focusing on, like a bottle of gas. Think of it as the “player” in a video game.
• Surroundings: Everything else around the system – the room, the table, even the air. It’s the “game world”.
• Internal Energy (U): The total energy stored inside the system’s particles (kinetic + potential). Imagine it as the amount of water stored in a tank.
• Heat (q): Energy transferred because of a temperature difference. Like water flowing from a higher pipe to a lower one.
• Work (w): Energy transferred when a force moves something. Picture pushing a shopping cart – the force you apply is work.

Mathematical Form of the First Law

When you rearrange the symbols you get the handy equation:

ΔU = q + w

That means:

• If you add 100 J of heat to a gas (q = +100 J) and the gas does 30 J of expansion work (w = –30 J), the internal energy rises by 70 J.
• If the system does 50 J of work on its surroundings and no heat is exchanged, ΔU = –50 J – the system loses energy.

Worked Example: Heating Water in a Closed Vessel

Suppose 200 g of water is sealed in a rigid container. We heat it so that its temperature rises from 25 °C to 75 °C. No volume change means no work (w = 0). The specific heat capacity of water is 4.18 J g⁻¹ °C⁻¹.

Calculate the change in internal energy.

1. Find the heat absorbed: q = m·c·ΔT = 200 g × 4.18 J g⁻¹ °C⁻¹ × (75‑25)°C = 200 × 4.18 × 50 = 41,800 J.
2. Since w = 0, ΔU = q + w = 41,800 J + 0 = 41,800 J.

So the water’s internal energy increased by about 42 kJ.

Types of Processes and Energy Changes

Different ways a system can exchange heat or work lead to distinct names. The table below sums them up.

Quick Summary (Bullet Version)

• Energy never disappears; it just moves or changes form.
• ΔU = q + w captures that balance for any chemical system.
• Positive q = heat added; negative q = heat lost.
• Positive w = work done on the system; negative w = work done by the system.
• Rigid container → w = 0, so ΔU equals the heat absorbed or released.

📝 Likely Exam Questions

1. State the first law of thermodynamics and explain each term.
   Answer: The first law states that the change in internal energy (ΔU) of a system equals the heat added to the system (q) plus the work done on the system (w). ΔU = q + w.
2. A 500 J of heat is supplied to a gas that does 150 J of work during expansion. Find ΔU.
   Answer: ΔU = q + w = 500 J + (–150 J) = 350 J.
3. Why does the internal energy of an ideal gas depend only on temperature?
   Answer: For an ideal gas, kinetic energy of particles (which determines internal energy) is directly proportional to temperature, and potential energy between particles is negligible.
4. Explain what happens to ΔU, q, and w in an isochoric process.
   Answer: In an isochoric (constant volume) process, work w = 0 because there’s no volume change. Thus ΔU = q; any heat added or removed changes internal energy directly.
5. Calculate the heat absorbed when 0.5 mol of a gas at 300 K is heated to 350 K at constant volume. (Cₚ = 29 J mol⁻¹ K⁻¹, Cᵥ = 20 J mol⁻¹ K⁻¹)
   Answer: q = n·Cᵥ·ΔT = 0.5 mol × 20 J mol⁻¹ K⁻¹ × (350‑300) K = 0.5 × 20 × 50 = 500 J.