Question

Calculate the change in entropy when 10.0 g of CO2 isothermally expands from a volume of 6.15 L to 11.5 L. Assume that the gas behaves ideally.

Answer 1

Answer:

The change in entropy of the carbon dioxide is $1.183\times 10^{-3}$ kilojoules per Kelvin.

Explanation:

By assuming that carbon dioxide behaves ideally, the change in entropy ($\Delta S$), measured in kilojoules per Kelvin, is defined by the following expression:

$\Delta S = m\cdot \bar c_{v}\cdot \ln \frac{T_{f}}{T_{o}}+m\cdot \frac{R_{u}}{M}\cdot \ln \frac{V_{f}}{V_{o}}$ (1)

Where:

$m$ - Mass of the gas, measured in kilograms.

$\bar c_{v}$ - Isochoric specific heat of the gas, measured in kilojoules per kilogram-Kelvin.

$T_{o}$, $T_{f}$ - Initial and final temperatures of the gas, measured in Kelvin.

$V_{o}$, $V_{f}$ - Initial and final volumes of the gas, measured in liters.

$R_{u}$ - Ideal gas constant, measured in kilopascal-cubic meter per kilomole-Kelvin.

$M$ - Molar mass, measured in kilograms per kilomole.

If we know that $T_{o} = T_{f}$, $m = 0.010\,kg$, $R_{u} = 8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K}$, $M = 44.010\,\frac{kg}{kmol}$, $V_{o} = 6.15\,L$ and $V_{f} = 11.5\,L$, then the change in entropy of the carbon dioxide is:

$\Delta S = \left[\frac{ (0.010\,kg)\cdot \left(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)}{44.010\,\frac{kg}{kmol} } \right]\cdot \ln \left(\frac{11.5\,L}{6.15\,L}\right)$

$\Delta S = 1.183\times 10^{-3}\,\frac{kJ}{K}$

The change in entropy of the carbon dioxide is $1.183\times 10^{-3}$ kilojoules per Kelvin.