The aim of this article is to find the **final temperature** of the gas after it has gone through a **polytropic process** of **compression** from **lower** to **higher pressure**.

The basic concept of this article is the **Polytropic process** and **Ideal Gas Law**.

The **polytropic process** is a **thermodynamic process** involving the **expansion** or **compression** of a gas resulting in **heat transfer**. It is expressed as follows:

\[PV^n\ =\ C\]

Where:

$P\ =$ **The pressure of the gas**

$V\ =$ **The volume of the gas**

$n\ =$ **Polytropic Index**

$C\ =$ **Constant**

## Expert Answer

Given that:

**Polytropic Index** $n\ =\ 1.2$

**Initial Pressure** $P_1\ =\ 120\ kPa$

**Initial Temperature** $T_1\ =\ 30°C$

**Final Pressure** $P_2\ =\ 1200\ kPa$

**Final Temperature** $T_2\ =\ ?$

First, we will convert the given temperature from **Celsius** to **Kelvin**.

\[K\ =\ ^{\circ}C+273\ =\ 30+273\ =\ 303K\]

Hence:

**Initial Temperature** $T_1\ =\ 303K$

We know that as per the** Polytropic process**:

\[PV^n\ =\ C\]

For a **polytropic process** between **two states**:

\[P_1{V_1}^n\ =\ P_2{V_2}^n\]

By rearranging the equation, we get:

\[\frac{P_2}{P_1}\ =\ \frac{{V_1}^n}{{V_2}^n}\ =\ \left(\frac{V_1}{V_2}\right)^n\]

As per **Idea Gas Law**:

\[PV\ =\ nRT\]

For **two states of gas**:

\[P_1V_1\ =\ nRT_{1\ }\]

\[V_1\ =\ \frac{nRT_{1\ }}{P_1}\]

And:

\[P_2V_2\ =\ nRT_2\]

\[V_2\ =\ \frac{nRT_2}{P_2}\]

Substituting the values from **Idea Gas law** into **Polytropic process relation**:

\[\frac{P_2}{P_1}\ =\ \left(\frac{\dfrac{nRT_{1\ }}{P_1}}{\dfrac{nRT_2}{P_2}}\right)^n\]

Cancelling $nR$ from **numerator** and **denominator**, we get:

\[\frac{P_2}{P_1}\ =\ \left(\frac{\dfrac{T_{1\ }}{P_1}}{\dfrac{T_2}{P_2}}\right)^n\]

\[\frac{P_2}{P_1}\ =\ \left(\frac{T_{1\ }}{P_1}\times\frac{P_2}{T_2}\right)^n\]

\[\frac{P_2}{P_1}\ =\ \left(\frac{P_{2\ }}{P_1}\times\frac{T_{1\ }}{T_2}\right)^n\]

\[\frac{P_2}{P_1}\ =\ \left(\frac{P_{2\ }}{P_1}\right)^n\times\left(\frac{T_{1\ }}{T_2}\right)^n\]

\[\left(\frac{T_{1\ }}{T_2}\right)^n\ =\ \left(\frac{P_{2\ }}{P_1}\right)^{1-n}\]

\[\frac{T_{1\ }}{T_2}\ =\ \left(\frac{P_{2\ }}{P_1}\right)^\dfrac{1-n}{n}\ or\ \ \frac{T_{2\ }}{T_1}\ =\ \left(\frac{P_{2\ }}{P_1}\right)^\dfrac{n-1}{n}\]

Now substituting the given values of **pressures** and **temperatures** of **argon gas** in **two states**, we get:

\[\frac{T_{2\ }}{303K}\ =\ \left(\frac{1200}{120}\right)^\dfrac{1.2-1}{1.2}\]

\[T_{2\ }\ =\ {303K\left(\frac{1200\ kPa}{120\ kPa}\right)}^\dfrac{1.2-1}{1.2}\]

\[T_{2\ }\ =\ {303K\times10}^{0.16667}\]

\[T_{2\ }\ =\ 444.74K\]

Converting the **Final Temperature** $T_{2\ }$ from **Kelvin** to **Celsius**, we get:

\[K\ =\ ^{\circ}C+273\]

\[444.74\ =\ ^{\circ}C+273\]

\[T_{2\ }\ =\ 444.74-273\ =171.74\ ^{\circ}C\]

## Numerical Result

The **Final Temperatur**e $T_{2\ }$ of the **argon gas** after it has gone through a **polytropic process** of **compression** from $120$ $kPa$ at $30^{\circ}C$ to $1200$ $kPa$ in a **piston-cylinder device**:

\[T_{2\ }=171.74\ ^{\circ}C\]

## Example

Determine the **final temperature** of **hydrogen gas** after it has gone through a **polytropic process** of **compression** with $n=1.5$ from $50$ $kPa$ and $80^{\circ}C$ to $1500$ $kPa$ in a **screw compressor**.

**Solution**

Given that:

**Polytropic Index** $n\ =\ 1.5$

**Initial Pressure** $P_1\ =\ 50\ kPa$

**Initial Temperature** $T_1\ =\ 80°C$

**Final Pressure** $P_2\ =\ 1500\ kPa$

**Final Temperature** $T_2\ =\ ?$

First, we will convert the given temperature from **Celsius** to **Kelvin**.

\[K\ =\ ^{\circ}C+273\ =\ 80+273\ =\ 353K\]

Hence:

**Initial Temperature** $T_1\ =\ 303K$

As per **polytropic process** expressions in term of **pressure** and **temperature**:

\[\frac{T_{2\ }}{T_1}\ =\ \left(\frac{P_{2\ }}{P_1}\right)^\dfrac{n-1}{n}\]

\[T_{2\ }\ =\ T_1\left(\frac{P_{2\ }}{P_1}\right)^\dfrac{n-1}{n}\]

Substituting the given values:

\[T_{2\ }\ =\ 353K\left(\frac{1500\ kPa}{50\ kPa}\right)^\dfrac{1.5-1}{1.5}\]

\[T_{2\ }\ =\ 353K\left(\frac{1500\ kPa}{50\ kPa}\right)^\dfrac{1.5-1}{1.5}\]

\[T_{2\ }\ =\ 1096.85K\]

Converting the **Final Temperature** $T_{2\ }$ from **Kelvin** to **Celsius**:

\[T_{2\ }\ =\ 1096.85-273\ =\ 823.85^{\circ}C \]