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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 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 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:

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

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