Calculator · Thermal Engineering
Ideal-Gas Internal Energy & Enthalpy
For an ideal gas internal energy and enthalpy depend on temperature only. With constant specific heats the changes are linear in ΔT, whatever process took the gas between the two temperatures.
Inputs
Dry air: cp ≈ 1004.5, cv ≈ 717.5.
Results
Δu = cv (T2 − T1)
71.75 kJ/kg
| T1 | 300 K |
|---|---|
| T2 | 400 K |
| ΔT | 100 K |
| Δu = cv ΔT | 71.75 kJ/kg |
| Δh = cp ΔT | 100.45 kJ/kg |
| ΔU = m Δu | 71.75 kJ |
| ΔH = m Δh | 100.45 kJ |
Assumptions
- · ideal gas
- · constant specific heats
Technical reference
Δu = cv (T2 − T1)
Δh = cp (T2 − T1)
ΔU = m Δu, ΔH = m Δh
- cv
- specific heat at constant volume[J/(kg·K)]
- cp
- specific heat at constant pressure[J/(kg·K)]
- T
- absolute temperature[K]
- m
- mass of gas[kg]
Specific heats are constant over the temperature range; over large ranges real gases need tabulated or temperature-dependent cp.
A temperature difference of 1 K equals 1 °C, but each temperature itself must be entered in kelvin.