Thermodynamics — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "Thermodynamics" — 8 important questions with detailed answers for CBSE board exam preparation.

TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Thermodynamics" — 8 important questions with detailed answers for CBSE board exam prep…

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Key Questions Covered:

  1. An ideal gas undergoes isothermal expansion from volume 1 L to 2 L at 300 K a…
  2. Calculate the work done on an ideal gas when it is compressed from 4 L to 1 L…
  3. One mole of an ideal gas at standard temperature and pressure (STP) undergoes…
  4. + 5 more questions in the full chapter

Solutions Summary:

Question Status
An ideal gas undergoes isothermal expansion from volume 1… ✓ Solved
Calculate the work done on an ideal gas when it is compre… ✓ Solved
One mole of an ideal gas at standard temperature and pres… ✓ Solved

Showing 3 of 8 questions

Q1: An ideal gas undergoes isothermal expansion from volume 1 L to 2 L at 300 K against a constant external pressure of 2 atm. Calculate the work done by the gas. (1 L·atm = 101.3 J)

Step 1: For isothermal expansion against constant external pressure: W = P_ext × ΔV Step 2: Calculate change in volume: ΔV = V_f - V_i = 2 - 1 = 1 L Step 3: Calculate work: W = 2 atm × 1 L = 2 L·atm Step 4: Convert to Joules: W = 2 × 101.3 = 202.6 J Alternatively, using W = nRT ln(V_f/V_i): For ...

Q2: Calculate the work done on an ideal gas when it is compressed from 4 L to 1 L at a constant pressure of 2 atm. (1 L·atm = 101.3 J)

Step 1: For compression at constant pressure (isobaric process): W = P × ΔV = P × (V_f - V_i) Step 2: Since volume decreases (compression), ΔV is negative: ΔV = V_f - V_i = 1 - 4 = -3 L Step 3: Calculate work done ON the gas: W = 2 atm × (-3 L) = -6 L·atm Step 4: Negative sign indicates work done...

Q3: One mole of an ideal gas at standard temperature and pressure (STP) undergoes adiabatic expansion to 10 times its original volume. If γ = 1.4, calculate the final temperature.

Step 1: For adiabatic process: TV^(γ-1) = constant T₁V₁^(γ-1) = T₂V₂^(γ-1) Step 2: Rearrange for T₂: T₂ = T₁ × (V₁/V₂)^(γ-1) Step 3: Substitute values: T₁ = 273 K (at STP) V₂/V₁ = 10, so V₁/V₂ = 0.1 γ = 1.4, so γ - 1 = 0.4 T₂ = 273 × (0.1)^0.4 Step 4: Calculate (0.1)^0.4: (0.1)^0.4 = (10⁻¹)^0.4 ...

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