# Argon is compressed in a polytropic process with n=1.2 from 120 kPa and 30°C to 1200 kPa in a piston-cylinder device. Determine the final temperature of the argon.

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 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

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 Temperature $T_{2\ }$ 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:

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