# Online Thermodynamics Calculator

Internal energy. The total energy of constituent molecules. It is equal to the sum of internal kinetic energy and potential energy. Change in internal energy dU = (U2 – U1) = nCvdT Internal energy is a function of state only and its change does not depend on path. First law of thermodynamics. The heat given is equal to sum of change in internal energy and work done by the system.

## Computational example

Isometric or isochoric change. Volume remains constant i.e., dV =0 δW = PdV = 0, δQ = dU = nCvdT. Isobaric change. Pressure remains constant i.e., dP =0 δQ = dU + δW = nCvdT + PdV. Isothermal change. Temperature remains constant i.e., dT =0 dU = nCvdT = 0, δQ = δW = PdV. Adiabatic change. No exchange of heat takes place i.e., δQ = 0 ∴ 0 = dU + δW or δW = – dU. Expansion in vacuum.

No transfer of heat and no work i.e., δQ = 0, δW = 0 ∴ dU = 0. Equation of state of gases. $$\frac{P V}{T}$$ = Constant PV = RT (for onle mole gas) PV = nRT (for n-mole gas). Isometric change: V = constant, $$\frac{P}{T}$$ = constant. Isobaric change: P = constant, $$\frac{V}{T}$$ = constant.

Adiabatic change: Entropy S = constant and ΔQ = 0 PVγ = constant TVγ-1 = constant P1-γTγ = constant Pd1-γ = constant. The slope of P-V curve –. Isothermal process $$\frac{\mathrm{dP}}{\mathrm{dV}}=-\frac{\mathrm{P}}{\mathrm{V}}$$. Adiabatic process $$\frac{\mathrm{dP}}{\mathrm{dV}}=-\gamma \frac{\mathrm{P}}{\mathrm{V}}$$. Work done by gases. W = $$\int_{V_{1}}^{V_{2}}$$PdV Isometric change: W = 0 Isobaric change: W = P(V2 – V1).

Isothermal change Wiso = nRTloge(V2/V1) = nRT 2.303 loge(V2/V1) = nRT loge(P1/P2).

Adiabatic change Wad = $$\frac{P_{1} V_{1}-P_{2} V_{2}}{\gamma-1}$$ = $$\frac{n R}{(\gamma-1)}$$(T1 – T2).

Elasticities of gases. Isothermal elasticity Eiso = P Adiabatic elasticity Eadia = γP. Take your subject knowledge to the next level by referring to the formulas of Physics provided by experts at Onlinecalculator.guru.

Reviewed byBogna Szyk. Combined gas law calculator is a great tool to deal with problems related to the most common transformations of gases.

Read about isobaric, isochoric, isothermal, and adiabatic processes of ideal gases (gases that can be described by the ideal gas equation).

And how it is possible for the ideal gases to do work or release/absorb heat. Check out the exact values for real gases and forget about struggling with thermodynamic exercises! An ideal gas can be described by several parameters, which are pressure p, volume V, temperature T and the amount of particles n.

They are correlated with the equation: p * V = n * R * T, where R stands for ideal gas constant and equals 8.3144598 J/(mol * K). During any process at least two of these properties change which can be compiled in combined gas law formula: p * V / T = k, where k is a constant.

SubstanceMass, kgSpecific heat, J/kg*CInitial temperature, CFinal temperature, C