Question

Consider the free, isothermal (constant T) expansion of an ideal gas. “Free” means that the external force is zero, perhaps because a stopcock has been opened and the gas is allowed to expand into a vacuum. Calculate $\mathbf{\Delta U}$for thisirreversible process. Show that q = 0, so that the expansion is also adiabatic$\left(q=0\right)$ for an ideal gas. This is analogous to a classic experiment first performed by Joule.

Short answer

q = 0

Step-by-step solution

Isothermal expansion of an ideal gas.

The isothermal process constitutes a variation in the system so that the temperature remains constant.

The collisions among the molecules are elastic in an ideal gas. There is no intermolecular attractive force.

The external pressure.

For an ideal gas,

$$\begin{gathered} {\text{∆U = nC}}_{\text{v}}\text{∆T} \\ \text{= 0} \end{gathered}$$

if T = 0 (isothermal), the value of${\text{P}}_{\text{ext}}$ is 0.

Show that q = 0, so the expansion is adiabatic for an ideal gas.

Here, the work can be found as:

$$\begin{gathered} {\text{W = -P}}_{\text{ext}}\text{∆V} \\ \text{= 0} \end{gathered}$$

The value of q is;

$$\begin{gathered} \text{q =}\text{U∆- w} \\ \text{= 0 - 0} \end{gathered}$$