Question

An isothermal process is one in which the temperatures of the system and surroundings remain constant at all times. With this in mind, what is wrong with the following statement: “For an isothermal expansion of an ideal gas against a constant pressure,$\mathbf{∆}\mathit{T}$ = 0, so q = 0”? What is qequal to in an isothermal expansion of an ideal gas against a constant external pressure?

Short answer

The q for an isothermal expansion of an ideal gas against a constant external pressure is $P_{ext}∆V$.

Step-by-step solution

Determination of q

When the gas is expanding against a constant pressure, the gas is doing work.

For isothermal process, and are zero.

$\mathbf{∆}\mathit{E}\mathbf{=}\mathit{q}\mathbf{+}\mathit{w}$

If w is not zero, then q will also be not zero.

0 = q + w

q = - w

Explanation of the given statement

For work against constant pressure, the following equation determine w,

$$\begin{gathered} \mathit{w}\mathbf{=}\mathbf{-}{\mathit{P}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{∆}\mathit{V} \\ \mathit{q}\mathbf{=}\mathbf{-}\mathbf{(}\mathbf{-}{\mathit{P}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{∆}\mathit{V}\mathbf{)} \\ \mathit{q}\mathbf{=}{\mathit{P}}_{\mathbf{e}\mathbf{x}\mathbf{t}}\mathbf{∆}\mathit{V} \end{gathered}$$