🔥 PV Diagram Thermodynamics Visualizer
ThermodynamicsWhat's happening? Work done = area under the P-V curve. Isothermal curves are steeper than adiabatic. Constant volume does zero work. This is the heart of JEE thermodynamics.
Thermodynamics is a high-weightage, concept-heavy topic — P-V diagrams and process identification are asked every year.
JEE tests isothermal, adiabatic, isobaric, and isochoric processes, work done from P-V area, and the first law ΔU = Q - W. The sign convention for work is the biggest source of errors.
Key formulas
Thermodynamics Formulas
First law
ΔU = Q - W
Change in internal energy = heat added - work done BY system. Sign convention: W > 0 when system expands.
Ideal gas law
PV = nRT
n = moles, R = 8.314 J/mol·K. For 1 mole: PV = RT.
Isothermal
PV = const
W = nRT·ln(V₂/V₁)
T constant. Work = area under curve = nRT·ln(Vf/Vi).
Adiabatic
PVγ = const
W = (P₁V₁ - P₂V₂)/(γ-1)
γ = Cp/Cv. No heat exchange (Q = 0).
Work done
W = ∫PdV
Area under P-V curve. Positive when volume increases (expansion), negative when volume decreases (compression).
Internal energy
ΔU = nCvΔT
Depends only on temperature change, not on the process. U = (3/2)nRT for monoatomic gas.
Step-by-step problem solving
How to solve thermodynamics problems in JEE
- Identify the process — isothermal (T const), adiabatic (Q=0), isobaric (P const), isochoric (V const).
- Write the relevant equation — PV = nRT, PVγ = const, or ΔU = nCvΔT.
- Calculate work — W = area under P-V curve. Sign: +W for expansion (V increases), -W for compression.
- Apply first law — ΔU = Q - W. Find Q or ΔU as required. Remember ΔU depends only on ΔT.
- Check γ values — monoatomic: γ = 5/3. Diatomic: γ = 7/5. Don't mix them up.
Common JEE mistakes
What students get wrong
❌ Wrong work sign
W = ∫PdV. If V increases (expansion), W > 0 (work done BY gas). If V decreases (compression), W < 0 (work done ON gas). Many students reverse this.
❌ Adiabatic vs isothermal
Isothermal: PV = const, T constant, heat flows. Adiabatic: PVγ = const, Q = 0, T changes. Easy to confuse the formulas.
❌ Assuming ΔU = 0 for all processes
ΔU = 0 only for isothermal processes (ΔT = 0). For adiabatic, ΔU = -W (since Q = 0). For all others, ΔU = nCvΔT.
❌ Wrong γ values
Monoatomic gas (monatomic = single atom like He, Ar): γ = 5/3, Cv = 3R/2. Diatomic (H₂, N₂, O₂): γ = 7/5, Cv = 5R/2. Don't swap them!
FAQs
Thermodynamics Questions
Is work done in a cycle always zero?
No — work done in a full cycle = area enclosed by the cycle on the P-V diagram. For a cyclic process, ΔU = 0 (returns to same state), so Q = W (net heat = net work done).
Why is adiabatic steeper than isothermal?
In an adiabatic process, temperature changes (no heat in/out), so pressure drops faster as volume increases. γ > 1, so PVγ = const drops more steeply than PV = const.
Can work be negative?
Yes — when the system compresses (volume decreases), work done BY the system is negative. You do work ON the system (compressing it), so W < 0. The gas loses energy.
Why does internal energy depend only on temperature?
For an ideal gas, there are no intermolecular forces — internal energy is purely kinetic (translational KE of molecules). KE depends on temperature only, not on P or V. So ΔU = nCvΔT always.