Question

Determine the change in entropy for 2.7 moles of an ideal gas originally placed in a container with a volume of 4.0 L when the container was expanded to a final volume of 6.0 L at constant temperature.

Answer 1

The value of entropy change for the process
`dS = 0.009 (KJ)/(K)`

Step-by-step explanation:

Mass of the ideal gas = 0.0027 kilo mol

Initial volume
`V_(1)` = 4 L

Final volume
`V_(2)` = 6 L

Gas constant for this ideal gas ( R ) =
`R_(u) M`

Where
`R_(u)` = Universal gas constant = 8.314
`(KJ)/(Kmol K)`

⇒ Gas constant R = 8.314 × 0.0027 = 0.0224
`(KJ)/(K)`

Entropy change at constant temperature is given by,

`dS = R log _(e) (V_(2))/(V_(1))`

Put all the values in above formula we get,

`dS = 0.0224 log _(e) [(6)/(4)]`

`dS = 0.009 (KJ)/(K)`

This is the value of entropy change for the process.