Work Done in Basic Thermodynamic Processes

Basic goal: determine dE, dQ, and dW for general thermodynamics processes. We now study 3 fundamental processes.

Constant Volume (Isochoric)

A constant volume process is the vertical path dV = 0 in the P-V plane---up if heat is added and down if heat is removed. Because dV = 0, the work done is dW = - P dV = 0

The First Law of Thermodynamics then states: dE= dQ + dW = dQ.

For a monatomic ideal gas: E =
3

2
NkT → dQ = dE =
3

2
Nk dT

Constant Pressure (Isobaric)

A constant pressure process is a horizontal path in the P-V diagram---right for expansion and left for compression. Example: a gas in a container sealed with a freely-sliding massive piston.

The work done during gas expansion is: dW = - P dV = - Nk dT

The First Law gives: dQ = dE - dW =
3

2
Nk dT + Nk dT =
5

2
Nk dT

Constant Temperature (Isothermal)

A constant temperature process is an isothermal path in the P-V diagram---a hyperbolic isotherm. Example: a gas in a container that is immersed in a constant-temperature bath is allowed to expand slowly, or is compressed slowly.

At constant temperature, the pressure of an ideal gas is: P = NkT/V.

The work done on the gas is:
W = - ∫_{v_i}^v_f P dV = - NkT ∫_{v_i}^v_f
1

V
dV

W = - NkT ln (
V_f

V_i
) = - P_i V_i ln (
V_f

V_i
) = - P_f V_f ln (
V_f

V_i
)

The First Law gives: dE = dQ + dW = 0 → dQ = - dW.

In an isothermal process, there is no change in the internal energy of an ideal gas.

Important: The work done during any thermodynamic process is path dependent (see transparency).