Isochoric process

Formalism

An isochoric thermodynamic quasi-static process is characterized by constant volume, i.e., ΔV = 0. The process does no pressure-volume work, since such work is defined by $${\displaystyle W=P\mathrm{\Delta }V,}$$ where P is pressure. The sign convention is such that positive work is performed by the system on the environment.

If the process is not quasi-static, the work can perhaps be done in a volume constant thermodynamic process.

For a reversible process, the first law of thermodynamics gives the change in the system's internal energy: $${\displaystyle dU=dQ-dW}$$

Replacing work with a change in volume gives $${\displaystyle dU=dQ-PdV}$$

Since the process is isochoric, dV = 0, the previous equation now gives $${\displaystyle dU=dQ}$$

Using the definition of specific heat capacity at constant volume, c_v = (dQ/dT)/m, where m is the mass of the gas, we get $${\displaystyle dQ=m{c}_{\mathrm{v}}dT}$$

Integrating both sides yields $${\displaystyle \mathrm{\Delta }Q=m{\int }_{T_1}^{T_2}{c}_{\mathrm{v}}dT,}$$ where c_v is the specific heat capacity at constant volume, T₁ is the initial temperature and T₂ is the final temperature. We conclude with: $${\displaystyle \mathrm{\Delta }Q=m{c}_{\mathrm{v}}\mathrm{\Delta }T}$$

On a pressure volume diagram, an isochoric process appears as a straight vertical line. Its thermodynamic conjugate, an isobaric process would appear as a straight horizontal line.

Ideal gas

If an ideal gas is used in an isochoric process, and the quantity of gas stays constant, then the increase in energy is proportional to an increase in temperature and pressure. For example a gas heated in a rigid container: the pressure and temperature of the gas will increase, but the volume will remain the same.