Isothermal or adiabatic: same gas, two paths
"Isothermal or adiabatic: same gas, two paths" is a free, interactive Class 11 Physics simulation for NEET and JEE, built on the lesson "Adiabatic process" in Thermodynamics. Pick the process, then slide the piston. Isothermal: the cylinder sits on a big reservoir at 300 K. Adiabatic: it is wrapped in insulation, so no heat can cross. Isothermal: PV = constant, ΔU = 0, all heat in becomes work out. Adiabatic: PV^γ = constant, Q = 0, all work comes from (or goes into) internal energy, so the adiabat is steeper. The work is the area under each curve. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Adiabatic process in Thermodynamics. Change the values and watch what happens.
In JEE Main
42 shifts analysed
48 questions from Thermodynamics across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §11.8.3
- In an adiabatic process the system is insulated, so no heat enters or leaves: ΔQ = 0. The first law then gives ΔW = −ΔU: work done by the gas comes out of its internal energy, so an ideal gas cools as it expands adiabatically.
- For an ideal gas undergoing a quasi-static adiabatic change, PV^γ = constant, where γ = C_p/C_v (ratio of the specific heats, ordinary or molar). The result is quoted without proof.
- So between two states: P₁V₁^γ = P₂V₂^γ. Since γ > 1, an adiabat is steeper than an isotherm through the same point: for the same compression the pressure rises more.
- Fig. 11.8: two adiabats connect two isotherms on a P–V diagram.
- Work done by the gas from (P₁, V₁, T₁) to (P₂, V₂, T₂): W = ∫P dV with P = constant/V^γ, giving W = [P₁V₁ − P₂V₂]/(γ − 1) = μR(T₁ − T₂)/(γ − 1).
- If the gas does work (W > 0), T₂ < T₁: it cools. If work is done on it (W < 0), T₂ > T₁: it warms up.
- Example: a gas with γ = 1.4 (such as hydrogen) compressed adiabatically to half its volume: P₂/P₁ = 2^1.4 ≈ 2.64, while an isothermal halving only doubles the pressure.
- For the same case, TV^(γ−1) = constant gives T₂ = T₁ × 2^0.4 ≈ 1.32 T₁: 300 K becomes about 396 K.
Questions
3 answers
- What does the "Isothermal or adiabatic: same gas, two paths" simulation show?
- Pick the process, then slide the piston. Isothermal: the cylinder sits on a big reservoir at 300 K. Adiabatic: it is wrapped in insulation, so no heat can cross.
- Does JEE Main ask questions from Thermodynamics?
- Yes. JEE Main asked 48 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Adiabatic process" in Thermodynamics (Class 11 Physics).
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