Question

Calculate the entropy change when 10 moles of an ideal gas expands reversibly and isothermally from an initial volume of 10L to 100L at 300K.

Answer 1

Hint: Entropy is the measure of randomness in a system. In an isothermal change the temperature of the system remains constant. To solve this numerical we will use the formula for change in entropy.
\[\Delta S\,=\,2.303nR\log \dfrac{{{V}_{2}}}{{{V}_{1}}}\]

Complete answer:
The concept of entropy was introduced by the second law of thermodynamics. The second law states that the entropy of the system becomes 0 at absolute 0K. The second law can be expressed in many forms. This law was brought because of the limitations of the first law.
The first law could not explain why reversible work of expansion for an ideal gas is greater than the reversible work of expansion of a real gas. It also does not tell us about the direction of change.
So, to account for all these limitations other laws of thermodynamics were brought.
Let’s look at the solution of the given numerical
First we will see the given information in the question:
Number of moles = 10
Initial volume of gas (${{V}_{1}}$) = 10L
Final volume of gas (${{V}_{2}}$) = 100L
Now, we will calculate the entropy of the gas using the formula:
\[\Delta S\,=\,2.303nR\log \dfrac{{{V}_{2}}}{{{V}_{1}}}\]
On putting the values, we get,
\[\Delta S\,=\,2.303\times 10\times 8.314\log \dfrac{100}{10}\]
\[\Delta S\,=\,191.47\,J{{K}^{-1}}\]
Therefore, the value of entropy for the given change in volume of the gas is 191.47J/K
Hence, the answer for the given question is $191.47\,J{{K}^{-1}}$

Additional Information:
Entropy is a state function. This means that the value of entropy doesn’t depend on the path of the change but only on the initial and final states.

Note:
In the above question students should use the formula of entropy according to the information given in the question. If the amount of heat is given then the formula used will change.
In that case the entropy will be calculated using the formula:
\[\Delta S\,=\,\dfrac{{{q}_{rev}}}{T}\]where, q is the heat change and T is the temperature.