Answer:
The change in entropy of the carbon dioxide is [tex]1.183\times 10^{-3}[/tex] kilojoules per Kelvin.
Explanation:
By assuming that carbon dioxide behaves ideally, the change in entropy ([tex]\Delta S[/tex]), measured in kilojoules per Kelvin, is defined by the following expression:
[tex]\Delta S = m\cdot \bar c_{v}\cdot \ln \frac{T_{f}}{T_{o}}+m\cdot \frac{R_{u}}{M}\cdot \ln \frac{V_{f}}{V_{o}}[/tex] (1)
Where:
[tex]m[/tex] - Mass of the gas, measured in kilograms.
[tex]\bar c_{v}[/tex] - Isochoric specific heat of the gas, measured in kilojoules per kilogram-Kelvin.
[tex]T_{o}[/tex], [tex]T_{f}[/tex] - Initial and final temperatures of the gas, measured in Kelvin.
[tex]V_{o}[/tex], [tex]V_{f}[/tex] - Initial and final volumes of the gas, measured in liters.
[tex]R_{u}[/tex] - Ideal gas constant, measured in kilopascal-cubic meter per kilomole-Kelvin.
[tex]M[/tex] - Molar mass, measured in kilograms per kilomole.
If we know that [tex]T_{o} = T_{f}[/tex], [tex]m = 0.010\,kg[/tex], [tex]R_{u} = 8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K}[/tex], [tex]M = 44.010\,\frac{kg}{kmol}[/tex], [tex]V_{o} = 6.15\,L[/tex] and [tex]V_{f} = 11.5\,L[/tex], then the change in entropy of the carbon dioxide is:
[tex]\Delta S = \left[\frac{ (0.010\,kg)\cdot \left(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)}{44.010\,\frac{kg}{kmol} } \right]\cdot \ln \left(\frac{11.5\,L}{6.15\,L}\right)[/tex]
[tex]\Delta S = 1.183\times 10^{-3}\,\frac{kJ}{K}[/tex]
The change in entropy of the carbon dioxide is [tex]1.183\times 10^{-3}[/tex] kilojoules per Kelvin.
