Question

Consider 1.00 mole of an ideal gas that is expanded isothermally at 25°C from 2.45 $\times$10^-2 atm to 2.45 $\times$10^-3 atm in the following three irreversible steps:

Step 1: from 2.45 $\times$10^-2atm to 9.87 $\times$10^-3 atm
Step 2: from 9.87 $\times$ 10^-3 atm to 4.97 $\times$10^-3 atm

Step 3: from 4.97 $\times$10^-3atm to 2.45 $\times$10^-3 atm

Calculate $\mathbf{q}\mathbf{,}\mathbf{w}\mathbf{,}\mathbf{∆}\mathbf{E}\mathbf{,}\mathbf{}\mathbf{∆}\mathbf{S}\mathbf{,}\mathbf{}\mathbf{∆}\mathbf{H}\mathbf{}\mathbf{and}\mathbf{}\mathbf{∆}\mathbf{G}$for each step andfor the overall process.

Short answer

Answer

The values of $\mathbf{q}\mathbf{,}\mathbf{}\mathbf{w}\mathbf{,}\mathbf{}\mathbf{∆}\mathbf{E}\mathbf{,}\mathbf{}\mathbf{∆}\mathbf{S}\mathbf{,}\mathbf{}\mathbf{∆}\mathbf{H}\mathbf{}\mathbf{and}\mathbf{}\mathbf{∆}\mathbf{G}\mathbf{}$for each step and for the overall process is tabulated below

	q	w	$∆$E	$∆$S	$∆$H	$∆$G
Step 1	1480	$-$1480 J	0	7.56$\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$	0	$-$2250 J
Step 2	1240	$-$1240J	0	5.77$\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$	0	$-$1720 J
Step 3	1246	$-$1246 J	0	5.81$\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$	0	$-$1730 J
Total	3966	$-$3966	0	19.14$\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$	0	$-$5700 J

Step-by-step solution

Introduction to the concept

The following formula can be used tocalculate the amount of work completed:

$$\begin{gathered} \mathbf{w}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{P}\mathbf{\times }\mathbf{∆}\mathbf{V} \\ \mathbf{∆}\mathbf{V}\mathbf{=}{\mathbf{V}}_{\mathbf{f}}\mathbf{-}{\mathbf{V}}_{\mathbf{i}}\mathbf{=}\mathbf{nRT}\mathbf{\times }\left(\frac{1}{P_f}-\frac{1}{P_i}\right) \end{gathered}$$

The total of heat and work is the internal energy

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

Theentropy changeis computed as follows:

$\mathbf{∆}\mathbf{S}\mathbf{=}\mathbf{nRIn}\mathbf{\times }\left(\frac{P_1}{P_2}\right)$

Changes in Gibbs free energy are also related to enthalpy and entropy in the following way:

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

where

$\mathbf{w}\mathbf{}\mathbf{=}$Workdone

$\mathbf{P}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}$ Pressure

$\mathbf{V}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}$Volume

$\mathbf{n}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}$Number of moles

$\mathbf{R}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}$Universal gas constant

$\mathbf{T}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}$Temperature

$\mathbf{∆}\mathbf{E}$= Change in energy

$\mathbf{q}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}$ Heat
$\mathbf{∆}\mathbf{S}\mathbf{}\mathbf{=}$Change in Entropy
$\mathbf{∆}\mathbf{H}\mathbf{}\mathbf{=}$Gibbs free energy change

$\mathbf{∆}\mathbf{G}\mathbf{}\mathbf{=}$change in enthalpy

Determination of step 1: from  2.45×10-2 atm to 9.87×10-3 at

From the given$\mathbf{∆}\mathbf{E}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{0}$, so that$\mathbf{∆}\mathbf{T}\mathbf{=}\mathbf{0}$

$$\begin{gathered} ∆\mathrm{V}=1·00\mathrm{mol}\times 0·08206\raisebox{1ex}{$\mathrm{L}\times \mathrm{atm}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}\times \mathrm{K}$}\right.\times 298\mathrm{K}\times \\ \left(\frac{1}{9.87\times {10}^{-3}\mathrm{atm}}\times \frac{1}{2.45\times {10}^{-2}\mathrm{atm}}\right) \end{gathered}$$

=1480 L

$$\begin{gathered} \mathbf{w}\mathbf{=}\mathbf{-}\mathbf{P}\mathbf{\times }\mathbf{∆}\mathbf{V} \\ \mathrm{w}=-9.87\times {10}^{-3}\mathrm{atm}\times 1480\mathrm{L} \\ =-14.6\mathrm{L}\times \mathrm{atm} \\ =-14.6\mathrm{L}\times \mathrm{atm}\times 101.3\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}$}\right.\mathrm{atm} \end{gathered}$$

= $\mathbf{-}\mathbf{}\mathbf{1480}\mathbf{}\mathbf{J}$

Thus,

$$\begin{gathered} \mathrm{q}=-\mathrm{w}=\mathbf{1480}\mathbf{J} \\ ∆\mathrm{S}=1.00\mathrm{mol}\times 8.314\times \raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}\mathrm{K}$}\right.\times \mathrm{In}\times \left(\frac{2.45\times {10}^{-2}\mathrm{atm}}{9.87\times {10}^{-2}\mathrm{atm}}\right) \end{gathered}$$

= $\mathbf{7}\mathbf{.}\mathbf{56}\raisebox{1ex}{$\mathbf{J}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{K}$}\right.$

$$\begin{gathered} ∆\mathrm{G}=0-298\mathrm{K}\times 7.56\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right. \\ =\mathbf{-}\mathbf{2250}\mathbf{}\mathbf{J} \end{gathered}$$

Determination of step 2: from  9.87×10-3 atm to 4.93 ×10-3 atm

From the given $\mathbf{∆}\mathbf{E}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{0}\mathbf{,}$so that$\mathbf{∆}\mathbf{T}\mathbf{=}\mathbf{0}$

$$\begin{gathered} ∆\mathrm{V}=1.00\mathrm{mol}\times 0.08206\raisebox{1ex}{$\mathrm{L}\times \mathrm{atm}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}\times \mathrm{K}$}\right.\times \\ 298\mathrm{K}\times \left(\frac{1}{4.93\times {10}^{-3}\mathrm{atm}}\times \frac{1}{9.87\times {10}^{-3}\mathrm{atm}}\right) \\ =2483\mathrm{L} \end{gathered}$$

localid="1649393014465" $\mathbf{w}\mathbf{=}\mathbf{-}\mathbf{P}\mathbf{\times }\mathbf{∆}\mathbf{v}$

localid="1649164511940" $$\begin{gathered} \mathrm{w}=-4.93\times {10}^{-3}\mathrm{atm}\times 2483\mathrm{L} \\ =-12.2\mathrm{L}\times \mathrm{atm} \\ =-12.2\mathrm{L}\times \mathrm{atm}\times 101.3\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}\times \mathrm{atm}$}\right. \end{gathered}$$

= $\mathbf{-}\mathbf{1240}\mathbf{}\mathbf{J}$

Thus,

$\mathrm{q}=-\mathrm{w}=$$\mathbf{1240}\mathbf{}\mathbf{J}$

$∆\mathrm{S}=1.00\mathrm{mol}\times 8.314\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}\times \mathrm{K}$}\right.\times \mathrm{In}\times \left(\frac{9.87\times {10}^{-3}\mathrm{atm}}{4.93\times {10}^{-3}\mathrm{atm}}\right)$

localid="1649393253246" $\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{77}\raisebox{1ex}{$\mathbf{J}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{K}$}\right.$

localid="1649165100496" $∆\mathrm{G}=0-298\mathrm{K}\times 5.77\times \raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$

=$\mathbf{}\mathbf{-}\mathbf{}\mathbf{1720}\mathbf{}\mathbf{J}$

Determination of step 3: from 4.93×10-3 atm to 2.45 ×10-3 atm 

From the given $\mathbf{∆}\mathbf{E}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{∆}\mathbf{H}\mathbf{=}\mathbf{0}$so that $\mathbf{∆}\mathbf{T}\mathbf{=}\mathbf{0}$

$$\begin{gathered} ∆\mathrm{V}=1.00\mathrm{mol}\times 0.08206\times \raisebox{1ex}{$\mathrm{L}\times \mathrm{atm}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}\times \mathrm{K}$}\right.\times \\ 298\mathrm{K}\times \left(\frac{1}{2.45\times {10}^{-3}\mathrm{atm}}-\frac{1}{4.93\times {10}^{-3}\mathrm{atm}}\right) \\ =5021\mathrm{L} \end{gathered}$$

$\mathbf{w}\mathbf{=}\mathbf{-}\mathbf{P}\mathbf{\times }\mathbf{∆}\mathbf{V}$

localid="1649166197602" $$\begin{gathered} \mathrm{w}=-2.45\times {10}^{-3}\mathrm{atm}\times 5021\mathrm{L} \\ =-12.3\mathrm{L}\times \mathrm{atm} \\ =-12.3\mathrm{L}\times \mathrm{atm}\times 101.3\times \raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{L}\times \mathrm{atm}$}\right. \end{gathered}$$

= localid="1649393723250" $\mathbf{-}\mathbf{1246}\mathbf{}\mathbf{J}$

localid="1649393791417" $\mathrm{q}=-\mathrm{w}=\mathbf{1246}\mathbf{}\mathbf{J}$

localid="1649393651266" $$\begin{gathered} ∆\mathrm{S}=1.00\mathrm{mol}\times 8.314\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}\times \mathrm{K}$}\right.\times \mathrm{In}\left(\frac{4.93\times {10}^{-3}\mathrm{atm}}{2.45\times {10}^{-3}\mathrm{atm}}\right) \\ =\mathbf{5}\mathbf{.}\mathbf{81}\raisebox{1ex}{$\mathbf{J}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{K}$}\right. \end{gathered}$$

$∆\mathrm{G}=0-298\mathrm{K}\times 5.81\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$

$\mathbf{=}\mathbf{1730}\mathbf{}\mathbf{J}$

Therefore the values are:

	q	w	$∆\mathrm{E}$	$∆\mathrm{S}$	$∆\mathrm{H}$	$∆\mathrm{G}$
Step 1	1480 J	$-$1480 J	0	$7.56\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$	0	$-2250\mathrm{J}$
Step 2	1240 J	$-1240\mathrm{J}$	0	5.77$\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$	0	$-1720\mathrm{J}$
Step 3	1246 J	$-1246\mathrm{J}$	0	5.81 $\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$	0	$-1730\mathrm{J}$
Total	3966 J	$-3966\mathrm{J}$	0	$19.14\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.$	0	$-5700\mathrm{J}$