Question

Constant Temperature (Isothermal) Process.

Answer 1

An isothermal process is shown in the figure below. It consists of a constant temperature reservoir at temperature (T₁) surrounding a piston-cylinder arrangement. Assume that a perfect gas is at any instant, at the temperature of the system (T₁), is contained inside the cylinder. At the thermal equilibrium state, the temperature of the system and the surroundings are the same. 

Hence, there is no transfer of heat across the boundary. If the piston now moves slightly downward, expansion of the gas takes place increasing its volume by (dV) and consequently the pressure and temperature of the system drop by an amount of (dP) and (dT) respectively. Therefore, heat will flow from the surroundings until the system reaches the original temperature (T₁). The isothermal process will be possible only when the process is quasi-static. 

The (P-V) diagram of the isothermal expansion process is shown in the figure below :

Applying the first law of thermodynamics:

Q - W = △U = mC_v (T₂ - T₁)

Since (T₂ = T₁) for an isothermal process

then :

\(Q=W=\int ^2_1\,PdV\,..(1)\)

For an isothermal process, from Boyles’s law, we have:

PV = C → P = C/V .....(2)

Substituting equation (1) in (2), we get

\(Q=W=C\int^2_1dV/V=C\,In\,(\frac{V_2}{V_1})\)

Since P₁V₁ = P₂V₂ = mRT = C

then :

\(Q=W=P_1V_1\,In(\frac{V_2}{V_1})\,...(3)\)

or

\(Q=W=mRT\,In(\frac{V_2}{V_1})\,...(4)\)