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.

• A. 19.15 $J{K}^{-1}mo{l}^{-1}$
• B. -29.15 $J{K}^{-1}mo{l}^{-1}$
• C. 4.7 $J{K}^{-1}mo{l}^{-1}$
• D. Zero

Answer 1

The correct option is D Zero

Change in entropy for a expansion process is given as:
$△S=n{C}_{v}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$ 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,
$T{V}^{\gamma -1}$= constant
$\Rightarrow T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
$\Rightarrow \frac{T_2}{T_1}=(\frac{V_1}{V_2}{)}^{\gamma -1}$

putting this in the equation 1,
$△S=n{C}_{v}ln(\frac{V_1}{V_2}{)}^{\gamma -1}+nRln\frac{V_2}{V_1}$
$△S=(\gamma -1)n{C}_{v}ln\left(\frac{V_1}{V_2}\right)+nRln\frac{V_2}{V_1}$

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

Using this,
$\Rightarrow △S=(\frac{C_p}{C_v}-1)n{C}_{v}ln\left(\frac{V_1}{V_2}\right)+nRln\frac{V_2}{V_1}$

Simplifying it out, we get,
$\Rightarrow △S=\left(\frac{R}{C_v}\right)n{C}_{v}ln\left(\frac{V_1}{V_2}\right)+nRln\frac{V_2}{V_1}$
$\Rightarrow △S=\left(R\right)nln\left(\frac{V_1}{V_2}\right)+nRln\frac{V_2}{V_1}$
$\Rightarrow △S=-\left(R\right)nln\left(\frac{V_2}{V_1}\right)+nRln\frac{V_2}{V_1}$
$\Rightarrow △S=0$


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Theory:

Calculation of $\Delta S_{universe}$ 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,
$\Delta S_{sys}=n{C}_{v,m}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$.....eq. 1
We know for reversible adiabatic process,
$T{V}^{\gamma -1}$= constant
$\Rightarrow T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
$\Rightarrow \frac{T_2}{T_1}=(\frac{V_1}{V_2}{)}^{\gamma -1}$

putting this in the equation 1,
$△S=n{C}_{v}ln(\frac{V_1}{V_2}{)}^{\gamma -1}+nRln\frac{V_2}{V_1}$
$△S=(\gamma -1)n{C}_{v}ln\left(\frac{V_1}{V_2}\right)+nRln\frac{V_2}{V_1}$

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

Using this,

$\Rightarrow △S=(\frac{C_p}{C_v}-1)n{C}_{v}ln\left(\frac{V_1}{V_2}\right)+nRln\frac{V_2}{V_1}$

Simplifying it out, we get,
$\Rightarrow △S=\left(\frac{R}{C_v}\right)n{C}_{v}ln\left(\frac{V_1}{V_2}\right)+nRln\frac{V_2}{V_1}$
$\Rightarrow △S=\left(R\right)nln\left(\frac{V_1}{V_2}\right)+nRln\frac{V_2}{V_1}$
$\Rightarrow △S=-\left(R\right)nln\left(\frac{V_2}{V_1}\right)+nRln\frac{V_2}{V_1}$
$\Rightarrow △S_{sys}=0$
and also $\Rightarrow △S_{surr}=0$


For a irreversible adiabatic process :

In adiabatic processes $\Delta S_{rev}$ is not equal to $\Delta S_{irr.}$. 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,
$\Delta S_{sys}=n{C}_{v,m}ln\frac{T_2}{T_1}+nRln\frac{V_2}{V_1}$
which is greater than zero.
and
$\Delta S_{surr}$ = 0
$\Delta S_{univ}=\Delta S_{sys}+\Delta S_{surr}$, which is greater than zero.
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