Select a thermodynamic process — isothermal, isobaric, isochoric, or adiabatic — then enter your gas parameters (pressure, volume, temperature, moles) to compute the final state variables, work done, heat transfer, and internal energy change for an ideal gas.
Work Done (W)
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Heat Transfer (Q)
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Internal Energy Change (ΔU)
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Final Pressure (p₂)
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Final Volume (V₂)
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Final Temperature (T₂)
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The four main thermodynamic processes for ideal gases are: <strong>isothermal</strong> (constant temperature, T = const), <strong>isobaric</strong> (constant pressure, p = const), <strong>isochoric</strong> (constant volume, V = const), and <strong>adiabatic</strong> (no heat exchange with surroundings, Q = 0). Each process constrains one state variable while the others change according to the ideal gas law.
In an isothermal process the temperature remains constant, so the internal energy of an ideal gas does not change (ΔU = 0) and all heat absorbed equals the work done: Q = W. Work is calculated as W = nRT·ln(V₂/V₁), where n is moles, R = 8.3145 J/(mol·K), T is the constant temperature, and V₁, V₂ are the initial and final volumes.
An isobaric process occurs at constant pressure. Work done by the gas is W = p·(V₂ − V₁), and because temperature changes, the internal energy also changes. Heat transfer is Q = n·Cp·(T₂ − T₁), where Cp is the molar heat capacity at constant pressure. The relationship between volumes and temperatures follows V₁/T₁ = V₂/T₂.
An isochoric (or isovolumetric) process keeps volume constant, so no work is done by or on the gas (W = 0). All heat added goes directly into changing the internal energy: Q = ΔU = n·Cv·(T₂ − T₁). Pressure and temperature are directly proportional: p₁/T₁ = p₂/T₂.
In an adiabatic process no heat is exchanged with the surroundings (Q = 0), so all work comes at the expense of internal energy: W = −ΔU. The relationship between state variables is p₁·V₁^γ = p₂·V₂^γ, where γ (gamma) is the heat capacity ratio Cp/Cv — equal to 1.4 for diatomic gases like N₂ and O₂, and 1.667 for monatomic gases like helium and argon.
Yes, but only during an <strong>isochoric (constant volume) process</strong>. Gay-Lussac's Law states p/T = constant when V is fixed, meaning doubling the temperature doubles the pressure. In other processes (isobaric, isothermal, adiabatic), pressure and temperature do not share this simple linear relationship.
The combined gas law is p₁V₁/T₁ = p₂V₂/T₂. Rearranging gives T₂ = T₁·(p₂·V₂)/(p₁·V₁). Simply plug in the known initial state (p₁, V₁, T₁) and the known final pressure p₂ and volume V₂ to find the final temperature T₂. Remember temperatures must be in Kelvin.
Use γ = 1.4 for diatomic gases such as nitrogen (N₂), oxygen (O₂), and air. Use γ = 1.667 for monatomic gases such as helium (He) and argon (Ar). Use γ ≈ 1.3 for triatomic gases such as carbon dioxide (CO₂) and water vapour. This ratio determines how much of the internal energy converts to work in an adiabatic process.