Thermodynamics Solved Examples (Class 11 Physics)
Thermodynamics numericals explore internal energy, heat transfer, work, and the first law of thermodynamics. These problems develop understanding of energy
TL;DR: Thermodynamics numericals explore internal energy, heat transfer, work, and the first law of thermodynamics. These problems develop understanding of e…
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Thermodynamics numericals explore internal energy, heat transfer, work, and the first law of thermodynamics. These problems develop understanding of energy
Thermodynamics — Solved Numerical Examples (Step by Step)
Example 1: 10 mol of an ideal gas expands from 1 L to 10 L at constant temperature 300 K. Calculate the work done by the gas. (R = 8.314 J/mol·K)
Solution: n = 10 mol, V₁ = 1 L = 0.001 m³, V₂ = 10 L = 0.01 m³, T = 300 K (constant). Work W = nRT ln(V₂/V₁) = 10 × 8.314 × 300 × ln(10) = 24942 × 2.303 = 57,402 J.
Example 2: 2 mol of ideal gas at constant pressure 100 kPa expands from 20 L to 40 L. Calculate work done and change in internal energy. (γ = 1.4)
Solution: Work at constant pressure W = P × ΔV = 100,000 × (0.04 - 0.02) = 100,000 × 0.02 = 2000 J. Using PV = nRT: T₁ = (100,000 × 0.02) / (2 × 8.314) = 1200 K. Using first law: Q = nCp × ΔT. For diatomic gas Cp = 3.5R. ΔT = 600 K, so Q = 2 × 3.5 × 8.314 × 600 = 69,888 J. ΔU = Q - W = 69,888 - 2000 = 67,888 J.
Example 3: A gas absorbs 500 J of heat. If 200 J of work is done by the gas, calculate the change in internal energy.
Solution: Using first law of thermodynamics: ΔU = Q - W. Where Q = heat absorbed = 500 J, W = work done by gas = 200 J. ΔU = 500 - 200 = 300 J.
Example 4: 1 mol of ideal gas undergoes adiabatic expansion from 1 atm to 0.5 atm. If initial temperature is 300 K, find final temperature. (γ = 1.4)
Solution: For adiabatic process: T₁ × P₁^(1-γ)/γ = T₂ × P₂^(1-γ)/γ. Or T₁ × P₁^((1-γ)/γ) = T₂ × P₂^((1-γ)/γ). Using T × P^((1-γ)/γ) = constant: 300 × 1^(-0.4/1.4) = T₂ × 0.5^(-0.4/1.4). 300 × 1 = T₂ × 0.5^(-0.286). T₂ = 300 / (0.821) = 365 K.
Example 5: Calculate the heat required to raise the temperature of 2 kg of water from 20°C to 100°C. (Specific heat of water = 4200 J/kg·°C)
Solution: Mass m = 2 kg, Initial temperature T₁ = 20°C, Final temperature T₂ = 100°C. ΔT = 80°C = 80 K. Heat Q = m × c × ΔT = 2 × 4200 × 80 = 672,000 J.
Tips
- First law: ΔU = Q - W (internal energy change equals heat absorbed minus work done).
- For adiabatic processes, Q = 0, so all work comes from internal energy decrease.
- Specific heat varies with the process: Cp (constant pressure) > Cv (constant volume).
Frequently Asked Questions
What is the difference between heat and temperature?
Temperature is the measure of kinetic energy of particles (how fast they move). Heat is the transfer of thermal energy from a hotter object to a cooler object.
Why do pressurized containers get hot when compressed?
In adiabatic compression, work is done on the gas, increasing its internal energy (since Q = 0). Increased internal energy means higher temperature.
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