Question

124. Consider the isothermal expansion of 1.00 mole of ideal gas at 27°C. The volume increases from 30.0 L to 40.0 L. Calculate$\mathbf{q}\mathbf{,}\mathbf{w}\mathbf{,}\mathbf{∆}\mathbf{E}\mathbf{,}\mathbf{∆}\mathbf{H}\mathbf{,}\mathbf{∆}\mathbf{S}\mathbf{}\mathbf{and}\mathbf{}\mathbf{∆}\mathbf{G}$ for two situations:
a. a free expansion
b. a reversible expansion

Short answer

1. For free expansion, the values are $$\begin{gathered} \mathbf{q}=0\mathrm{J} \\ \mathbf{w}=0\mathrm{J} \\ \mathbf{∆}\mathbf{E}=0\mathrm{J} \\ \mathbf{∆}\mathbf{H}=0\mathrm{J} \\ \mathbf{∆}\mathbf{S}=2.39\mathrm{J}/\mathrm{K} \\ \mathbf{∆}\mathbf{G}=-717\mathrm{J} \end{gathered}$$
2. For reversible expansion, the values are $$\begin{gathered} \mathbf{q}=718 \\ \mathbf{w}=-718 \\ \mathbf{∆}\mathbf{E}=0\mathrm{J} \\ \mathbf{∆}\mathbf{H}=0\mathrm{J} \\ \mathbf{∆}\mathbf{S}=2.39\mathrm{J}/\mathrm{K} \\ \mathbf{∆}\mathbf{G}=-717\mathrm{J} \end{gathered}$$

Step-by-step solution

Step 1: Introduction to the Concept

Isothermal expansion occurs when a substance expands in volume at a constant temperature.

Step 2: Determination of q,w,∆E,∆H,∆S and ∆G by free expansion

a)

There isno heat exchange with the surroundings in an isothermal process for an ideal gas, hence are$\mathbf{∆}\mathbf{E}\mathbf{}\mathbf{and}\mathbf{}\mathbf{∆}\mathbf{H}$bothzero.

There isno work done because this is a free expansion against a vacuum.

For anisothermal process, the following equation can be used to get the value of q.

$$\begin{gathered} ∆\mathrm{E}=\mathrm{q}+\mathrm{w} \\ \mathrm{q}=∆\mathrm{E}-\mathrm{w} \\ \mathrm{q}=0 \end{gathered}$$

The following equation can be used tocompute the change in entropy.

$\mathbf{∆}\mathbf{S}\mathbf{=}\mathbf{nRln}\left(\frac{V_2}{V_1}\right)$

$$\begin{gathered} ∆\mathrm{S}=\left(1.00\mathrm{mole}\right)\times \left(8.314\mathrm{J}/\mathrm{Kmol}\right)\times \ln \left(\frac{40.0\mathrm{L}}{30.0\mathrm{L}}\right) \\ ∆\mathrm{S}=2.39\mathrm{J}/\mathrm{K} \end{gathered}$$

Calculate the change in free energy using these numbers.

$\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{∆}\mathbf{H}\mathbf{-}\mathbf{T}\mathbf{∆}\mathbf{S}$

$$\begin{gathered} ∆\mathrm{G}=0-\left(300\mathrm{K}\right)\times \left(2.39\mathrm{J}/\mathrm{K}\right) \\ ∆\mathrm{G}=-717\mathrm{J} \end{gathered}$$

Therefore

$$\begin{gathered} \mathbf{q}=0\mathrm{J} \\ \mathbf{w}=0\mathrm{J} \\ \mathbf{∆}\mathbf{E}=0\mathrm{J} \\ \mathbf{∆}\mathbf{H}=0\mathrm{J} \\ \mathbf{∆}\mathbf{S}=2.39\mathrm{J}/\mathrm{K} \\ \mathbf{∆}\mathbf{G}=-717\mathrm{J} \end{gathered}$$

Step 3: Determination of q,w,∆E,∆H,∆S and ∆G by reversible expansion

b)

There is no heat exchange with the environment in an isothermal process with an ideal gas, hence $\mathbf{∆}\mathbf{E}\mathbf{}\mathbf{and}\mathbf{}\mathbf{∆}\mathbf{H}$both zero.

The following formula calculates the work required for a reversible expansion:

$\mathbf{w}\mathbf{=}\mathbf{-}\mathbf{nRTln}\left(\frac{V_2}{V_1}\right)$

$$\begin{gathered} \mathrm{w}=-\left(1.00\mathrm{mole}\right)\times \left(8.314\mathrm{J}/\mathrm{Kmole}\right)\times \left(300\mathrm{K}\right)\times \ln \left(\frac{40.0\mathrm{L}}{30.0\mathrm{L}}\right) \\ \mathrm{w}=-718\mathrm{J} \end{gathered}$$

For an isothermal process, the following equation can be used to getthe value of q.

$$\begin{gathered} \mathrm{q}=-\mathrm{w} \\ \mathrm{q}=718\mathrm{J} \end{gathered}$$

The following equation can be used to compute thechange in entropy:

$\mathbf{∆}\mathbf{S}\mathbf{=}\mathbf{nRln}\left(\frac{V_2}{V_1}\right)$

$$\begin{gathered} ∆\mathrm{S}=\left(1.00\mathrm{mole}\right)\times \left(8.314\mathrm{J}/\mathrm{Kmol}\right)\times \ln \left(\frac{40.0\mathrm{L}}{30.0\mathrm{L}}\right) \\ ∆\mathrm{S}=2.39\mathrm{J}/\mathrm{K} \end{gathered}$$

Calculate the change in free energy using these numbers.

$\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{∆}\mathbf{H}\mathbf{-}\mathbf{T}\mathbf{∆}\mathbf{S}$

$$\begin{gathered} ∆\mathrm{G}=0-\left(300.0\mathrm{K}\right)\times \left(2.39\mathrm{J}/\mathrm{K}\right) \\ ∆\mathrm{G}=-717\mathrm{J} \end{gathered}$$

Therefore

$$\begin{gathered} \mathbf{q}=718\mathrm{J} \\ \mathbf{w}=-718\mathrm{J} \\ \mathbf{∆}\mathbf{E}=0\mathrm{J} \\ \mathbf{∆}\mathbf{H}=0\mathrm{J} \\ \mathbf{∆}\mathbf{S}=2.39\mathrm{J}/\mathrm{K} \\ \mathbf{∆}\mathbf{G}=-717\mathrm{J} \end{gathered}$$