249,433.88
Pa
498,867.76
Pa
300
K
2
20.7862
J/mol·K
6,235.85
J
6,235.85
J
0
J
249,433.88
Pa
498,867.76
Pa
300
K
2
20.7862
J/mol·K
6,235.85
J
6,235.85
J
0
J
The Isochoric Process Calculator (also known as an isometric or constant-volume process calculator) analyzes thermodynamic changes that occur within a rigid, fixed-volume container. Since the volume does not change, the gas performs no boundary work ($$W = 0$$), and all heat added to or removed from the system goes entirely into changing the internal energy.
The relationship $$Q = \Delta U = nC_v\Delta T$$ makes isochoric processes the simplest to analyze thermodynamically. The pressure and temperature are linked by Gay-Lussac's law: $$P_1/T_1 = P_2/T_2$$. This means heating a gas in a sealed container causes the pressure to rise proportionally to the absolute temperature — a critical consideration for pressure vessel safety.
Isochoric processes appear naturally in rigid containers such as sealed pressure vessels, bomb calorimeters, and the constant-volume heating phase of the Otto cycle (spark-ignition engines). In a bomb calorimeter, the reaction vessel is rigid, so the measured heat equals the internal energy change of the reaction, providing direct access to $$\Delta U$$ rather than $$\Delta H$$.
This calculator determines initial and final pressures using the ideal gas law, verifies that work is zero, and computes the heat transferred and internal energy change. The molar heat capacity at constant volume $$C_v = R/(\gamma - 1)$$ depends on the gas type through the heat capacity ratio $$\gamma$$, allowing analysis of monatomic, diatomic, and polyatomic gases.
The calculator uses the ideal gas law and constant-volume thermodynamic relations:
Pressures from ideal gas law: $$P = \frac{nRT}{V}$$, giving $$P_1 = \frac{nRT_1}{V}$$ and $$P_2 = \frac{nRT_2}{V}$$
Gay-Lussac's Law: $$\frac{P_2}{P_1} = \frac{T_2}{T_1}$$ (at constant volume)
Work: $$W = \int P\,dV = 0$$ (volume is constant, $$dV = 0$$)
Internal energy change: $$\Delta U = nC_v\Delta T = n\frac{R}{\gamma - 1}(T_2 - T_1)$$
First law: $$Q = \Delta U + W = \Delta U$$ (since $$W = 0$$)
All heat input goes directly into raising the gas temperature and internal energy. No energy escapes as mechanical work, making isochoric heating the most direct way to increase a gas's thermal energy.
A positive heat value means heat flows into the gas — temperature and pressure both rise. A negative heat value means the gas cools — temperature and pressure both drop. The work is always exactly zero because the rigid container prevents expansion. The pressure ratio equals the temperature ratio, so doubling the absolute temperature doubles the absolute pressure. Engineers must design pressure vessels to withstand the maximum expected pressure after heating, with appropriate safety factors.
Inputs
Results
1 mol in 10 L heated from 300 K to 600 K. Pressure doubles from ~249 kPa to ~499 kPa. All 6236 J of heat goes into internal energy with no work done.
Inputs
Results
2 mol of monatomic gas (γ = 5/3) cooled from 500 K to 250 K. Pressure halves and 6236 J of heat is rejected from the system.
An isochoric (or isometric) process is a thermodynamic process at constant volume. The term comes from Greek: iso (equal) and chora (space/volume). Since the volume doesn't change, no boundary work is done ($$W = 0$$), and all heat transferred equals the change in internal energy.
Boundary work is defined as $$W = \int P\,dV$$. When volume is constant, $$dV = 0$$, so the integral evaluates to zero regardless of pressure. Physically, the rigid walls prevent the gas from expanding, so no displacement occurs against external pressure and no mechanical work is performed.
According to Gay-Lussac's law, pressure is directly proportional to absolute temperature at constant volume: $$P \propto T$$, or $$P_1/T_1 = P_2/T_2$$. Doubling the absolute temperature doubles the pressure. This is why sealed containers can become dangerously pressurized when heated.
$$C_v$$ is the molar heat capacity at constant volume — the heat needed to raise 1 mol by 1 K without volume change. $$C_p$$ is the heat capacity at constant pressure. For ideal gases, $$C_p - C_v = R$$. At constant volume, all heat goes to internal energy; at constant pressure, some heat also does expansion work, requiring more total heat.
The Otto cycle (gasoline engines) models combustion as isochoric heat addition — the spark ignites the fuel-air mixture so rapidly that the piston barely moves, approximating constant volume. The exhaust blowdown is also modeled as isochoric heat rejection. This distinguishes the Otto cycle from the Diesel cycle, which uses isobaric heat addition.
A bomb calorimeter is a thick-walled sealed container used to measure the heat of combustion of a sample. Because the vessel is rigid and sealed, the reaction occurs at constant volume. The measured heat equals $$\Delta U$$ (internal energy change) rather than $$\Delta H$$ (enthalpy change), though the two are related by $$\Delta H = \Delta U + \Delta(nRT)$$.
Roboculator Team
The Roboculator Team explains calculations, planning tools, and practical formulas in clear language for real-life situations.
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