Thermodynamics: A Comprehensive Guide for Class 11

Fundamental Laws: Zeroth and First Law of Thermodynamics

Thermodynamics is built upon a set of fundamental laws. The first one we encounter, perhaps counter-intuitively, is the Zeroth Law. It states: If two systems are each in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. This might sound obvious, but it's the very principle that allows us to use a thermometer! The thermometer (system C) reaches equilibrium with your body (system A) and gives a reading. If it gives the same reading for another object (system B), the Zeroth Law tells us that your body and the object are at the same temperature.

The First Law of Thermodynamics is essentially the law of conservation of energy applied to a thermodynamic system. It introduces three key quantities: Internal Energy (U), the sum of all kinetic and potential energies of the molecules in a system; Heat (Q), energy transferred due to a temperature difference; and Work (W), energy transferred when a force acts over a distance (like a gas expanding against a piston). The law is mathematically stated as:

ΔQ = ΔU + ΔW

This equation means that the heat (ΔQ) supplied to a system is used for two purposes: to increase its internal energy (ΔU) and to do work (ΔW) on its surroundings.

Understanding Key Thermodynamic Processes

Isothermal Process

A process where the temperature of the system remains constant (ΔT = 0). For an ideal gas, this means the internal energy also remains constant (ΔU = 0). The First Law becomes ΔQ = ΔW.

Adiabatic Process

A process where there is no exchange of heat between the system and its surroundings (ΔQ = 0). The process must happen very quickly or in a perfectly insulated container. The First Law becomes 0 = ΔU + ΔW, or ΔW = -ΔU. Work is done at the expense of internal energy.

Isochoric Process

A process where the volume of the system remains constant (ΔV = 0). Since no volume change occurs, the work done is zero (ΔW = 0). The First Law simplifies to ΔQ = ΔU. All heat supplied goes into increasing the internal energy.

Isobaric Process

A process where the pressure of the system remains constant (ΔP = 0). Work is done as the volume changes, calculated as W = PΔV. The First Law remains in its full form: ΔQ = ΔU + PΔV.

Worked Examples: Applying the First Law

• Example 1: Isochoric Heating A rigid container holds 2 moles of a monatomic ideal gas. 500 J of heat is supplied to the gas. What is the change in its internal energy? Step 1: Identify the process. The container is 'rigid', which means its volume is constant. This is an isochoric process. Step 2: Recall the work done in an isochoric process. In an isochoric process, ΔV = 0. The work done, W = PΔV, is therefore zero. W = 0 J. Step 3: Apply the First Law of Thermodynamics. The First Law is ΔQ = ΔU + ΔW. Step 4: Substitute the known values. We are given ΔQ = +500 J (heat is supplied) and we found W = 0 J. So, 500 J = ΔU + 0. Final Answer: The change in the internal energy of the gas is ΔU = 500 J.
• Example 2: Isobaric Expansion An ideal gas expands from a volume of 2 m³ to 4 m³ at a constant pressure of 100 kPa. If 500 kJ of heat is supplied to the gas, calculate the change in its internal energy. Step 1: Identify the process and variables. The process occurs at 'constant pressure', so it is isobaric. Initial volume V₁ = 2 m³ Final volume V₂ = 4 m³ Pressure P = 100 kPa = 100 × 10³ Pa Heat supplied ΔQ = 500 kJ = 500 × 10³ J Step 2: Calculate the work done by the gas. For an isobaric process, the work done by the gas is W = PΔV = P(V₂ - V₁). W = (100 × 10³ Pa) × (4 m³ - 2 m³) W = (100 × 10³ Pa) × (2 m³) = 200 × 10³ J = 200 kJ. Step 3: Apply the First Law of Thermodynamics. The First Law is ΔQ = ΔU + ΔW. We need to find ΔU. Step 4: Rearrange and solve for ΔU. ΔU = ΔQ - ΔW ΔU = (500 × 10³ J) - (200 × 10³ J) = 300 × 10³ J. Final Answer: The change in internal energy is 300 kJ.

The Second Law of Thermodynamics and Heat Engines

While the First Law tells us that energy is conserved, it doesn't tell us the direction in which processes occur. For instance, a cup of hot coffee always cools down in a room; it never spontaneously gets hotter by drawing heat from the cooler room. The Second Law of Thermodynamics addresses this directionality. It can be stated in several ways:

• Kelvin-Planck Statement: It is impossible to construct a device that operates in a cycle and produces no other effect than the extraction of heat from a single body and the performance of an equivalent amount of work. This means no heat engine can be 100% efficient.
• Clausius Statement: It is impossible to construct a device that operates in a cycle and produces no other effect than the transfer of heat from a colder body to a hotter body.

A heat engine is a practical device that embodies these principles. It takes heat (Q₁) from a high-temperature source (a 'hot reservoir' at T₁), converts part of it into useful work (W), and rejects the remaining heat (Q₂) to a low-temperature sink (a 'cold reservoir' at T₂). By conservation of energy, W = Q₁ - Q₂. The efficiency (η) of a heat engine is the ratio of the work done to the heat absorbed from the source: η = W / Q₁ = (Q₁ - Q₂) / Q₁ = 1 - (Q₂/Q₁). The Second Law guarantees that Q₂ is always greater than zero, so the efficiency η is always less than 1 (or 100%).

Exam Tip: Master the Sign Conventions

One of the most common sources of error in thermodynamics problems is incorrect sign conventions for heat (Q) and work (W). Always use the standard physics convention based on the system's perspective:

• Heat (Q):
• Q > 0: Heat is added to the system.
• Q < 0: Heat is removed from the system.

• Work (W):
• W > 0: Work is done by the system (e.g., gas expansion).
• W < 0: Work is done on the system (e.g., gas compression).

Remember the First Law equation: ΔQ = ΔU + ΔW. If a gas is compressed (W < 0) adiabatically (ΔQ = 0), then 0 = ΔU + W, which means ΔU = -W. Since W is negative, ΔU becomes positive, and the internal energy (and temperature) increases. Getting the signs right is crucial for a correct answer!

Practice Questions with Solutions

• Q: A gas in a cylinder is compressed from a volume of 5.0 L to 2.0 L by a constant external pressure of 150 kPa. During this process, 400 J of heat is released by the gas. Find the change in internal energy of the gas. A: Step 1: Identify the given values and their signs. The process is isobaric compression. P = 150 kPa = 1.5 × 10⁵ Pa ΔV = V_final - V_initial = 2.0 L - 5.0 L = -3.0 L = -3.0 × 10⁻³ m³ Heat is released, so ΔQ = -400 J. Step 2: Calculate the work done on the gas. W = PΔV = (1.5 × 10⁵ Pa) × (-3.0 × 10⁻³ m³) = -450 J. The negative sign correctly indicates that work is done ON the system. Step 3: Apply the First Law of Thermodynamics: ΔQ = ΔU + ΔW. -400 J = ΔU + (-450 J) Step 4: Solve for ΔU. ΔU = -400 J + 450 J = 50 J. Final answer: The change in internal energy of the gas is +50 J.
• Q: During an adiabatic expansion, a gas does 100 J of work. What is the change in its internal energy? A: Step 1: Identify the process. The process is adiabatic. Step 2: Recall the condition for an adiabatic process. In an adiabatic process, there is no heat exchange with the surroundings, so ΔQ = 0. Step 3: State the work done. The gas does work, so the work done BY the system is W = +100 J. Step 4: Apply the First Law of Thermodynamics: ΔQ = ΔU + ΔW. 0 = ΔU + 100 J Step 5: Solve for ΔU. ΔU = -100 J. Final answer: The change in internal energy is -100 J. The energy to do the work came from the internal energy of the gas, causing it to decrease.
• Q: A heat engine absorbs 2000 J of heat from a source at 500 K and rejects 1200 J of heat to a sink. What is the efficiency of the engine? A: Step 1: Identify the heat absorbed and heat rejected. Heat absorbed from the source, Q₁ = 2000 J. Heat rejected to the sink, Q₂ = 1200 J. Step 2: Calculate the work done by the engine. The work done is the difference between the heat absorbed and the heat rejected. W = Q₁ - Q₂ = 2000 J - 1200 J = 800 J. Step 3: Calculate the efficiency (η). Efficiency is the ratio of work done to the heat absorbed. η = W / Q₁ η = 800 J / 2000 J = 0.4. Step 4: Express the efficiency as a percentage. Efficiency = 0.4 × 100% = 40%. Final answer: The efficiency of the heat engine is 40%.
• Q: An ideal gas undergoes a cyclic process ABCA. The work done during AB is 50 J, during BC is 0 J, and 30 J of heat is rejected by the gas during CA. What is the work done during the process CA? A: Step 1: Understand the properties of a cyclic process. In a cyclic process, the system returns to its initial state. Therefore, the net change in internal energy (ΔU_net) is zero. Step 2: Apply the First Law to the entire cycle. ΔQ_net = ΔU_net + ΔW_net. Since ΔU_net = 0, we have ΔQ_net = ΔW_net. Step 3: Calculate the net work done (ΔW_net). ΔW_net = W_AB + W_BC + W_CA. We are given W_AB = 50 J and W_BC = 0 J. So, ΔW_net = 50 J + 0 J + W_CA. Step 4: Calculate the net heat exchange (ΔQ_net). We know heat is rejected during CA, so Q_CA = -30 J. The problem does not give heat information for AB or BC, but it gives us enough to relate net work and net heat. The question is slightly tricky; it asks for W_CA using heat information. A key insight for cyclic processes is that net work equals net heat. Thus, the work done during a part of the cycle is related to the heat in other parts. Let's re-read. Ah, let's assume Q_AB and Q_BC are also involved. The net heat is Q_net = Q_AB + Q_BC + Q_CA. The net work is W_net = W_AB + W_BC + W_CA = 50 + 0 + W_CA. Since Q_net = W_net, we have Q_AB + Q_BC - 30 = 50 + W_CA. This has too many unknowns. Let's rethink the problem statement. Correction: Usually, in such problems, you are given net heat or enough information to find it. Let's assume a simpler version often asked: If the net heat absorbed in the cycle is 20J, find W_CA. Let's solve that version: Q_net = 20 J. W_net = W_AB + W_BC + W_CA = 50 J + 0 J + W_CA. Since Q_net = W_net, 20 J = 50 J + W_CA. W_CA = 20 J - 50 J = -30 J. Let's solve the original question as stated, which is a common setup. For a cyclic process, ΔU_net = 0. So ΔQ_net = ΔW_net. ΔQ_net = Q_AB + Q_BC + Q_CA and ΔW_net = W_AB + W_BC + W_CA. The question implies we should find W_CA from the given data alone. For a full cycle, W_net = Area enclosed by the P-V diagram. Q_net is the net heat absorbed. Let's assume the question meant '30 J of work is done on the gas during CA'. If W_CA = -30 J. Then the net work is W_net = 50 + 0 - 30 = 20 J. The net heat absorbed would be 20 J. The original question seems incomplete, but the most logical interpretation is that the missing heat term (e.g., Q_AB) must be found. Let's assume the question should have been: In a cyclic process ABCA, the change in internal energy is zero. If 80 J of heat is absorbed and the work done during AB and BC are 50 J and 0 J respectively, find the work done during CA. Solution to revised problem: ΔQ_net = +80 J. ΔW_net = W_AB + W_BC + W_CA = 50 + 0 + W_CA. For a cycle, ΔQ_net = ΔW_net. So, 80 J = 50 J + W_CA. W_CA = 30 J. Final answer (based on a logical interpretation): Without complete heat data, the question is ambiguous. However, based on typical problems, if net heat absorbed was, for example, 20J, the work done during CA would be -30J.

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