Question

Question: (a) If$\mathbf{2}\mathbf{.60}\mathbf{}{\mathbf{\text{molO}}}_{\mathbf{2}}\mathbf{(}\mathbf{}\mathbf{\text{g}}\mathbf{)}\left(c_{\text{P}}=29.4\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$is compressed reversibly and adiabatically from an initial pressure of $\mathbf{1}.\mathbf{00}\mathbf{atm}$and 300K to a final pressure of $8.\mathbf{00}\mathbf{atm}$, calculate ΔS for the gas.

(b) Suppose a different path from that in part (a) is used. The gas is first heated at constant pressure to the same final temperature, and then compressed reversibly and isothermally to the same final pressure. CalculateΔSfor this path and show that it is equal to that found in part (a).

Short answer

a) The total entropy change $\Delta S=0$.

b) The final temperature $T_f=537.7\text{K}$.

This path's overall entropy change is equivalent to that observed in part (a).

Step-by-step solution

Given data

$2.60{\text{molO}}_2(\text{g})\left(c_{\text{P}}=29.4\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$, Initial pressure $\mathbf{1}.\mathbf{00}\mathbf{atm}$, Final pressure $8.\mathbf{00}\mathbf{atm}$, Initial temperature 300K.

Concept used

Entropy Change is the phenomenon that is the measure of change of disorder or randomness in a thermodynamic system. It is related to the conversion of heat or enthalpy in work. The sum of the system and the surroundings is the total entropy change.

Calculate the final temperature of the process

The final temperature has been calculated as follows:

$\gamma =\frac{C_p}{C_v}$

$C_p$- the molar heat capacity at constant$P$.

$C_v$- the molar heat capacity at constant$T$.

$\begin{array}{l}{C}_V=C_P-R\\ \gamma =\frac{C_p}{C_P-R}\end{array}$

$\begin{array}{c}\gamma =\frac{29.4{\text{JK}}^{-1}{\text{mol}}^{-1}}{29.4{\text{JK}}^{-1}{\text{mol}}^{-1}-8.314{\text{JK}}^{-1}{\text{mol}}^{-1}}\\ =1.39\end{array}$

$P_i^{1-\gamma }{T}_i^{\gamma }=P_f^{1-\gamma }{T}_f^{\gamma }$the relation between temperature and pressure in adiabatic process

$P_i$and $P_f$- the initial and final pressures

$T_i$ and $T_f$- the initial and final temperatures

γ - the ratio of molar heat capacities

$T_f^{\gamma }=\frac{P_i^{1-\gamma }{T}_i^{\gamma }}{P_f^{1-\gamma }}$

$\begin{array}{c}{T}_f^{1.39}=\frac{{(1\text{atm})}^{-0.39}\times {(300\text{K})}^{1.39}}{{(8\text{atm})}^{-0.39}}\\ =6243.15{\text{K}}^{139}\\ T_f={\left(6243.15{\text{K}}^{1.39}\right)}^{\frac{1}{139}}\\ =537.7\text{K}\end{array}$

Calculate the total entropy change

In the different path, the gas is heated to 167.4K at constant P.

$\Delta S_1=n{C}_P\ln \left(\frac{T_f}{T_i}\right)$ - Entropy change for isobaric process.

After the isobaric expansion, the system underwent isothermal reversible compression to the final pressure of 8atm.

• Entropy change for this process as follows:

$\begin{array}{c}\Delta S_1=2.60\text{mol}\times 29.4{\text{JK}}^{-1}{\text{mol}}^{-1}\times \ln \left(\frac{537.7\text{K}}{300\text{K}}\right)\\ =44.60{\text{JK}}^{-1}\\ \Delta S_2=2.60\text{mol}\times 8.314{\text{JK}}^{-1}{\text{mol}}^{-1}\times \ln \left(\frac{1atm}{8atm}\right)\\ =-44.95{\text{JK}}^{-1}\end{array}$

• The total change of entropy of the process as follows:

$$\begin{gathered} \Delta S=\Delta S_1+\Delta S_2 \\ \begin{array}{c}\Delta S=44.60{\text{JK}}^{-1}+\left(-44.95{\text{JK}}^{-1}\right)\\ =-0.35{\text{JK}}^{-1}\\ =0\end{array} \end{gathered}$$

The total entropy change in this path is equal to that found in part (a)