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

Answer:

The value of entropy change for the process $dS = 0.009 \frac{KJ}{K}$

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 $\frac{KJ}{Kmol K}$

⇒ Gas constant R = 8.314 × 0.0027 = 0.0224 $\frac{KJ}{K}$

Entropy change at constant temperature is given by,

$dS = R log _{e} \frac{V_{2}}{V_{1}}$

Put all the values in above formula we get,

$dS = 0.0224 log _{e} [\frac{6}{4}]$

$dS = 0.009 \frac{KJ}{K}$

This is the value of entropy change for the process.