Entropy

What is Entropy?

Entropy is a measure of the disorder of the molecular motion of a system. i.e Greater is the disorder, greater is the entropy.

The change in entropy is given as

$d S=\frac{\text { Heat absorbed by system }}{\text { Absolute temperature }}$ or $d S=\frac{d Q}{T}$

The relation $d S=\frac{d Q}{T}$ is called the mathematical form of the Second Law of Thermodynamics

1. Entropy for solid and liquid

i. When heat is given to a substance to change its state at a constant temperature.

Then change in entropy is given as

$d S=\frac{d Q}{T}= \pm \frac{m L}{T}$

where the positive sign refers to heat absorption and the negative sign to heat evolution.

And $L=$ Latent Heat and T is in Kelvin.

ii. When heat is given to a substance to raise its temperature from $T_1$ to $T_2$

Then change in entropy is given as

$d S=\int \frac{d Q}{T}=\int_{T_1}^{T_2} m c \frac{d T}{T}=m c \log _e\left(\frac{T_2}{T_1}\right)=2.303 * \operatorname{mc} \log _{10}\left(\frac{T_2}{T_1}\right)$

where c = specific heat capacity

2. Entropy for an ideal gas

For n mole of an ideal gas, the equation is given as PV = nRT

I.Entropy change for ideal gas in terms of T & V

From the first law of thermodynamics, we know that $d Q=d W+d U$
and
$
\Delta S=\int \frac{d Q}{T}=\int \frac{n C_V d T+P d V}{T}
$
using $P V=n R T$
$
\begin{aligned}
& \Delta S=\int \frac{n C_V d T+\frac{n R T}{V} d V}{T}=n C_V \int_{T_1}^{T_2} \frac{d T}{T}+n R \int_{V_1}^{V_2} \frac{d V}{V} \\
& \Delta S=n C_V \ln \left(\frac{T_2}{T_1}\right)+n R \ln \left(\frac{V_2}{V_1}\right)
\end{aligned}
$
II. Entropy change for an ideal gas in terms of $T$ \& $P$
$
\Delta S=n C_P \ln \left(\frac{T_2}{T_1}\right)-n R \ln \left(\frac{P_2}{P_1}\right)
$

III. Entropy change for an ideal gas in terms of P \& V
$
\Delta S=n C_V \ln \left(\frac{P_2}{P_1}\right)+n C_P \ln \left(\frac{V_2}{V_1}\right)
$

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Solved Examples Based on Entropy

Example 1: Which of the following is incorrect regarding the first law of thermodynamics?

1) It introduces the concept of the internal energy

2) It introduces the concept of entropy

3) It is applicable to any cyclic process

4) It is a restatement of the principle of conservation of energy

Solution:

The first law of Thermodynamics

Heat imported to a body is in general used to increase internal energy and work done against external pressure.

wherein

$
d Q=d U+d W
$

Entropy
It is a measure of the disorder of molecular motion of a system.
wherein
Greater is disorder greater is entropy
$
d S=\frac{d Q}{T}
$

The concept of entropy is introduced in the second law of thermodynamics.

The first law dealt with internal energy, work, and heat energy.

It is a statement of the first law of thermodynamics.

Hence, the answer is the option (2).

Example 2: The temperature­- entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is

1) $1 / 3$
2) $2 / 3$
3) $1 / 2$
4) $1 / 4$

Solution:

Entropy

It is a measure of disorder of molecular motion of a system.

wherein

Greater is disorder greater is entropy

and

Efficiency of a cyclic process
$
\eta=\frac{\text { work done per cyclic }}{\text { gross heat supplied per cyclic }}
$
wherein
Gross heat supplied implies only part of heat absorbed.
The efficiency of cycle is:
$
\begin{aligned}
& \eta=\frac{\text { work done }}{\text { heat absorb }} \\
& =\frac{Q_1-Q_2}{Q_1} \\
& =\frac{\text { Area of cycle }}{\text { Area under AB curve }} \\
& \qquad \eta=\frac{\frac{1}{2} \times T_0 \times S_0}{T_0 S_0+\frac{1}{2} T_0 S_0}=\frac{1}{3} \\
& \text { }
\end{aligned}
$

Example 3: If $\Delta Q$ heat is given to a substance and changes its state at constant temperature T then a change in entropy ds will be
1) $\pm \frac{m L}{T}$
2) $\pm \frac{m C}{T}$
3) $\pm m L$
4) $\pm m C$

Solution:

Entropy for solid and liquid

$
\begin{aligned}
& \Delta S=\frac{m L}{T} \\
& L=\text { Latent Heat } \\
& \text { T in kelvin } \\
& \frac{d \theta}{d T}= \pm \frac{m L}{T}
\end{aligned}
$

When a substance changes its state at a constant temperature $d \theta= \pm m L$ where the (+ve) sign refers to heat absorption and the (-ve) sign to heat evolution

Hence, the answer is the option (1).

Example 4: When heat is given to a substance of mass $m$ and specific heat c its temperature raise from $27^{\circ} \mathrm{C}$ to $327^{\circ} \mathrm{C}$, then
1) $2.303 \log _{10}(2)$
2) $2.303 \log _{10}(1 / 2)$
3) $\log _{10}(3 / 2)$
4) $2.303 \log _{10}(3 / 2)$

Solution:

Entropy for solid and liquid

When heat is given to a substance to raise its temperature from $T_1$ to $T_2$
wherein
$
\begin{aligned}
& \Delta S=m s \ln \left(\frac{T_2}{T_1}\right) \\
& \mathrm{s}=\text { specific heat capacity } \\
& \mathrm{ds}=2.303 \\
& \log \left[\frac{273+327}{273+27}\right]=2.303 \log _{10} 2=0.693
\end{aligned}
$

Hence, the answer is the option 1,

Example 5: The change in the entropy of 1 mole of an ideal gas that went through an isothermal process from the initial position to the final state is equal to

${ }_{1)} R \ln \frac{V_2}{V_1}$
2) $R \ln \frac{V_1}{V_2}$
3) zero
4) $R \ln T$

Solution:

Entropy change for an ideal gas in terms of T & V

$
\Delta S=n c_p \ln \left(\frac{T_2}{T_1}\right)+n R \ln \left(\frac{V_2}{V_1}\right)
$

For an ideal gas, we have
$
\Delta S=n c_p \ln \left(\frac{T_2}{T_1}\right)+n R \ln \left(\frac{V_2}{V_1}\right)
$
for isothermal process
$
\begin{aligned}
& \log _e\left(\frac{T_2}{T_1}\right)=0 \\
& \Delta s=n R \log _e\left(\frac{V_2}{V_1}\right) \text { for one mole of gas } \\
& \text { So } \\
& \Delta s=R \log _e\left(\frac{V_2}{V_1}\right)
\end{aligned}
$

Hence, the answer is the option 1.

Summary

Entropy, or the shift in the energy state accompanied by heat, is connected to a number of daily actions. It aids in a shift in heat transport and is reflected by the first, second, and third laws of thermodynamics. The more unpredictability in the system, the higher the rate of entropy; conversely, the lower the randomness of the system's molecules, the lower the entropy.