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

The change in entropy of the carbon dioxide is
`1.183* 10^(-3)` kilojoules per Kelvin.

Step-by-step 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 (T_(f))/(T_(o))+m\cdot (R_(u))/(M)\cdot \ln (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\,(kPa\cdot m^(3))/(kmol\cdot K)`,
`M = 44.010\,(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[( (0.010\,kg)\cdot \left(8.315\,(kPa\cdot m^(3))/(kmol\cdot K) \right))/(44.010\,(kg)/(kmol) ) \right]\cdot \ln \left((11.5\,L)/(6.15\,L)\right)`

`\Delta S = 1.183* 10^(-3)\,(kJ)/(K)`

The change in entropy of the carbon dioxide is
`1.183* 10^(-3)` kilojoules per Kelvin.