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Entropy

1 mole of an ...

Question

1 mole of an ideal gas at $25^{\circ} C$ is subjected to expand reversibly and adiabatically to ten times of its initial volume. Calculate the change in entropy during expansion.

19.15

J K^{- 1} m o l^{- 1}

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-29.15

J K^{- 1} m o l^{- 1}

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4.7

J K^{- 1} m o l^{- 1}

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Zero

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Solution

The correct option is D Zero
Change in entropy for a expansion process is given as:
$△ S = n C_{v} l n \frac{T_{2}}{T_{1}} + n R l n \frac{V_{2}}{V_{1}}$ eq...1
where $T_{2}, V_{2}$ are the final temperature and volume respectively and $T_{1}, V_{1}$ is the initial temperature and volume.

We know for adiabatic process,
$T V^{γ - 1}$ = constant
$\Rightarrow T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
$\Rightarrow \frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{γ - 1}$

putting this in the equation 1,
$△ S = n C_{v} l n (\frac{V_{1}}{V_{2}})^{γ - 1} + n R l n \frac{V_{2}}{V_{1}}$
$△ S = (γ - 1) n C_{v} l n (\frac{V_{1}}{V_{2}}) + n R l n \frac{V_{2}}{V_{1}}$

We know that , $γ - 1 = \frac{C_{p . m}}{C_{v . m}} - 1 = \frac{R}{C_{v . m}}$

Using this,
$\Rightarrow △ S = (\frac{C_{p}}{C_{v}} - 1) n C_{v} l n (\frac{V_{1}}{V_{2}}) + n R l n \frac{V_{2}}{V_{1}}$

Simplifying it out, we get,
$\Rightarrow △ S = (\frac{R}{C_{v}}) n C_{v} l n (\frac{V_{1}}{V_{2}}) + n R l n \frac{V_{2}}{V_{1}}$
$\Rightarrow △ S = (R) n l n (\frac{V_{1}}{V_{2}}) + n R l n \frac{V_{2}}{V_{1}}$
$\Rightarrow △ S = - (R) n l n (\frac{V_{2}}{V_{1}}) + n R l n \frac{V_{2}}{V_{1}}$
$\Rightarrow △ S = 0$

Theory:

Calculation of $Δ S_{u n i v e r s e}$ for an adiabatic Process:

For a reversible adiabatic process :
We have to calculate the change in entropy of the system in adiabatic conditions when it is moving from state A to state B.

Change in entropy for system is given as,
$Δ S_{s y s} = n C_{v, m} l n \frac{T_{2}}{T_{1}} + n R l n \frac{V_{2}}{V_{1}}$ .....eq. 1
We know for reversible adiabatic process,
$T V^{γ - 1}$ = constant
$\Rightarrow T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1}$
$\Rightarrow \frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{γ - 1}$

putting this in the equation 1,
$△ S = n C_{v} l n (\frac{V_{1}}{V_{2}})^{γ - 1} + n R l n \frac{V_{2}}{V_{1}}$
$△ S = (γ - 1) n C_{v} l n (\frac{V_{1}}{V_{2}}) + n R l n \frac{V_{2}}{V_{1}}$

We know that , $γ - 1 = \frac{C_{p . m}}{C_{v . m}} - 1 = \frac{R}{C_{v . m}}$

Using this,

$\Rightarrow △ S = (\frac{C_{p}}{C_{v}} - 1) n C_{v} l n (\frac{V_{1}}{V_{2}}) + n R l n \frac{V_{2}}{V_{1}}$

Simplifying it out, we get,
$\Rightarrow △ S = (\frac{R}{C_{v}}) n C_{v} l n (\frac{V_{1}}{V_{2}}) + n R l n \frac{V_{2}}{V_{1}}$
$\Rightarrow △ S = (R) n l n (\frac{V_{1}}{V_{2}}) + n R l n \frac{V_{2}}{V_{1}}$
$\Rightarrow △ S = - (R) n l n (\frac{V_{2}}{V_{1}}) + n R l n \frac{V_{2}}{V_{1}}$
$\Rightarrow △ S_{s y s} = 0$
and also $\Rightarrow △ S_{s u r r} = 0$

For a irreversible adiabatic process :

In adiabatic processes $Δ S_{r e v}$ is not equal to $Δ S_{i r r .}$ . When we start the process from state A, the system ends in system B via an irreversible pathway and another system say system C via a reversible pathway. Also $T_{2}$ irr. is greater than $T_{1}$ rev.

Change in entropy for system in irreversible process is given as,
$Δ S_{s y s} = n C_{v, m} l n \frac{T_{2}}{T_{1}} + n R l n \frac{V_{2}}{V_{1}}$
which is greater than zero.
and
$Δ S_{s u r r}$ = 0
$Δ S_{u n i v} = Δ S_{s y s} + Δ S_{s u r r}$ , which is greater than zero.

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