Polytropic Process

A polytropic process is a thermodynamic process that follows the equation $P V^n=$ constant, where P is pressure, V is volume, and n is the polytropic index. This type of process is versatile, representing various specific thermodynamic processes such as isothermal, adiabatic, and isobaric, depending on the value of n. In real life, polytropic processes can be observed in systems like air compressors, internal combustion engines, and even in the natural cooling of gases in the atmosphere. For instance, the compression of air in a car engine cylinder during the intake stroke closely resembles a polytropic process, where the heat generated during compression is partially transferred to the surroundings, leading to an intermediate behaviour between adiabatic and isothermal processes. Understanding polytropic processes is essential in designing efficient engines, refrigeration systems, and other technologies that rely on the controlled manipulation of gases.

Polytropic Process

A polytropic process is a thermodynamic process that involves the relationship between pressure, volume, and temperature of a gas, described by the equation $P V^n=$ constant, where P is pressure, V is volume, and n is the polytropic index. This index determines the specific nature of the process, encompassing various well-known thermodynamic processes such as isothermal (constant temperature), adiabatic (no heat exchange), and isobaric (constant pressure).

A process $P V^N=C$ is called a polytropic process. So, any process in this world related to thermodynamics can be explained by a polytropic process.

For example - 1. If $\mathrm{N}=1$, then the process becomes isothermal.
2. If $\mathrm{N}=0$, then the process becomes isobaric.
3. If $\mathrm{N}=\gamma$, then the process become adiabatic

Work Done by the Polytropic Process

The work done by a gas during a polytropic process is an essential concept in thermodynamics. It represents the amount of energy transferred by the system as it expands or contracts.

$W_{1-2}=\int P d V$

For a polytropic process,
$
\begin{gathered}
P V^N=P_1 V_1^N=P_2 V_2^N=C \\
P=\frac{C}{V^N}
\end{gathered}
$

Substituting in Equation, we get,
$
\begin{aligned}
\int P d V & =\int \frac{C d V}{V^N}=C \int V^{-N} d v \\
& =\left[V^{1-N}\right]_1^2=\left(V_2^{1-N}-V_1^{1-N}\right) \\
W_{1-2} & =\frac{P_2 V_2-P_1 V_1}{1-N} \text { or } \frac{P_1 V_1-P_2 V_2}{N-1} \ldots \ldots(1) \\
P_1 V_1 & =n R T_1 \\
P_2 V_2 & =n R T_2
\end{aligned}
$

So, equation (1) can be written as

$
W_{1-2}=\frac{n R\left(T_2-T_1\right)}{1-N}
$

And for one mole, $W_{1-2}=\frac{R\left(T_2-T_1\right)}{1-N}$

Specific Heat for Polytropic Process

We can write the equation of heat as $Q=C \Delta T$

Here C = Molar specific heat

From the first law of thermodynamics
$
\begin{aligned}
& Q=\Delta U+W \\
& \text { or } C \Delta T=C_v \Delta T-\frac{R \Delta T}{(N-1)} \\
& \therefore \quad C=C_v-\frac{R}{(N-1)}=\frac{R}{(\gamma-1)}-\frac{R}{(N-1)} \\
&
\end{aligned}
$

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Solved Examples Based on Polytropic Process

Example 1: In a process, the temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT=K, where K is a constant. In this process, the temperature of the gas is increased by $\Delta T$. If the amount of heat absorbed by gas is $x \times R \Delta T$, then what will be the value of 'x' (R is gas constant ) :

1) 0.50

2) 0.66

3) 0.33

4) 1.33

Solution:

$\begin{aligned} & \mathrm{VT}=\mathrm{K} \\ & \mathrm{V}\left[\frac{\mathrm{PV}}{\mathrm{nR}}\right]=\mathrm{K} \\ & \mathrm{PV}^2=\mathrm{nRK} \\ & \because \mathrm{V}^2=\mathrm{K} \\ & \because \mathrm{C}=\frac{\mathrm{R}}{1-\mathrm{x}}+\mathrm{C}_{\mathrm{v}} \\ & \quad \text { ( for polytropic process ) } \\ & \mathrm{C}=\frac{\mathrm{R}}{1-2}+\frac{3 \mathrm{R}}{2}=\frac{\mathrm{R}}{2} \\ & \therefore \Delta \mathrm{Q}=\mathrm{nC} \Delta \mathrm{T}=\frac{\mathrm{R}}{2} \Delta \mathrm{T}\end{aligned}$

Therefore 'x' will be 0.5

Hence, the answer is the option (1).

Example 2: The work done by the 1 mole of N₂ gas undergoing process PV^1.2 during which its temperature changes from $27^{\circ} \mathrm{C}$ to $227^{\circ} \mathrm{C}$ is: (work done in joules )

1) 8314

2) -8314

3) -4157

4) 4157

Solution:

$
W_{1-2}=\frac{n R\left(T_1-T_2\right)}{N-1}
$

Given $\mathrm{N}=1.2$, Gas constant $\mathrm{R}=8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}$,
$
\begin{gathered}
T_1=27^{\circ} \mathrm{C} \text { and } T_2=227^{\circ} \mathrm{C} \\
W_{1-2}=\frac{1 \times 8.314(27-227)}{1.2-1}=-8314 \mathrm{~J}
\end{gathered}$

Hence, the answer is the option (2).

Example 3: In a certain thermodynamical process, the pressure of a gas depends on the volume $k V^3$. The work done when the temperature changes from $100^{\circ} \mathrm{C}$ to $300^{\circ} \mathrm{C}$ will be __________nR, where n denotes the number of moles of a gas.

1) 50

2) 60

3) 70

4) 80

Solution:

$\begin{aligned} & T_i=100^{\circ} \mathrm{C} \& T_{\mathrm{f}}=300^{\circ} \mathrm{C} \\ & \Delta \mathrm{T}=300-100 \\ & \Delta \mathrm{T}=200^{\circ} \mathrm{C} \\ & P=\mathrm{kV}^3 \\ & \text { now } \mathrm{PV}=\mathrm{nRT} \\ & \therefore \mathrm{kV}^4=\mathrm{nRT} \\ & \text { now } 4 \mathrm{kV} \mathrm{V}^3 \mathrm{dV}=\mathrm{nRdT} \\ & \therefore \mathrm{PdV}=\mathrm{nRdT} / 4 \\ & \therefore \text { Work }=\int \mathrm{PdV}=\int \frac{\mathrm{nRdT}}{4}=\frac{\mathrm{nR}}{4} \Delta \mathrm{T} \\ & \Rightarrow \frac{200}{4} \times \mathrm{nR}=50 \mathrm{nR}\end{aligned}$

Hence, the answer is the option (1).

Example 4: The thermodynamic process is shown below on a $P-V$ diagram for one mole of an ideal gas. If $V_2=2 V_1$ then the ratio of temperature $\frac{T_2}{T_1}$ is :

1) $\frac{1}{2}$
2) 2
3) $\sqrt{2}$
4) $\frac{1}{\sqrt{2}}$

Solution:

$\begin{aligned} & \mathrm{PV}^{1 / 2}=\mathrm{c} \\ & \mathrm{Using} P V=n R T \\ & \Rightarrow \frac{n \mathrm{RT}}{\mathrm{V}} * \mathrm{~V}^{1 / 2}=\mathrm{c} \\ & \Rightarrow \mathrm{T}=\mathrm{c}^1 \mathrm{~V}^{1 / 2} \\ & \Rightarrow \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\left(\frac{\mathrm{V}_2}{\mathrm{~V}_1}\right)^{1 / 2}=\left(\frac{2 \mathrm{~V}_1}{\mathrm{~V}_1}\right)^{1 / 2} \\ & \Rightarrow \frac{\mathrm{T}_2}{\mathrm{~T}_1}=\sqrt{2}\end{aligned}$

Hence, the answer is the option (3).

Example 5: The volume V of a given mass of monoatomic gas changes with temperature T according to the relation $V=K T^{\frac{2}{3}}$. The work done when the temperature changes by 90K will be xR. The value of x is _______

[R = universal gas constant]

1) 60

2) 270

3) 90

4) 45

Solution:

$
\begin{aligned}
& \mathrm{W}=\int \mathrm{PdV} \\
& \Rightarrow \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}} \\
& \Rightarrow \mathrm{W}=\int \frac{\mathrm{nRT}}{\mathrm{V}} \mathrm{dV} \ldots \\
& \text { and } \mathrm{V}=\mathrm{KT}^{2 / 3} \ldots \\
& \Rightarrow \mathrm{W}=\int \frac{\mathrm{nRT}}{\mathrm{KT}^{2 / 3}} \cdot \mathrm{dV} \ldots \\
& \Rightarrow \text { from }(4): \mathrm{dV}=\frac{2}{3} \mathrm{~K} \mathrm{~T}^{-1 / 3} \mathrm{~d} \mathrm{~T} \\
& \Rightarrow W=\int_{T_1}^{T_2} \frac{n R T}{K T^{2 / 3}} \frac{2}{3} K \frac{1}{T^{1 / 3}} d T \\
& \Rightarrow \mathrm{W}=\frac{2}{3} \mathrm{nR} \times\left(\mathrm{T}_2-\mathrm{T}_1\right) \ldots \\
& \Rightarrow \mathrm{T}_2-\mathrm{T}_1=90 \mathrm{~K} \\
& \Rightarrow W=\frac{2}{3} n R \times 90 \\
& \Rightarrow \mathrm{W}=60 \mathrm{nR}
\end{aligned}
$

Assuming 1 mole of gas
$
n=1
$

So $\mathrm{W}=60 \mathrm{R}$

Hence, the answer is the option (2).

Summary

A polytropic process is a versatile thermodynamic process described by $P V^n=$ constant, where $n$ determines the nature of the process, including isothermal, adiabatic, and isobaric. The work done during this process can be calculated using specific formulas depending on $n$, and it is crucial for understanding energy transfer in systems like engines and compressors. The concept also extends to calculating specific heat and solving practical problems involving changes in temperature, pressure, and volume in gases.