This problem aims to familiarize us with different **state laws** of **physics** and **chemistry** involving **temperature, volume,** and **pressure.** The concepts required to solve this problem include **Boyle’s** **law,** the** ideal gas law,** and** work done** using** polytropic processes.**

First, we will look at **Boyle’s law**, which is a **practical ****gas** **law** that defines how the **stress of gas molecules** on the walls of a cylinder manages to drop as the **volume** of the cylinder rises. Whereas the** ideal gas law** describes the visible **properties** of **ideal** gases.

Here, the phrase** polytropic** is used to express any **reversible** method. Such a process revolves around any** empty or sealed** system of **gas** or vapor. This applies to both **heat and work** transfer mechanisms, keeping in view that the **aforementioned properties** are kept **constant** throughout the procedure.

## Expert Answer

The **Formulas** required for this problem are:

\[ P_1 \times V^{n}_1 = P_2 \times V^{n}_2 \]

\[ W = \dfrac{P_2 \times V_2 – P_1 \times V_1}{1-n}\]

\[ m = \dfrac{P_1 \times V_1}{R\times T_1} \]

From the **statement,** we are given the following information:

The** initial Volume**, $V_1 = 0.07 m^3$.

The** initial Pressure**, $P_1 = 130 kPa$.

The** final Pressure**, $P_2 = 80 kPa$.

Now we will find the **final volume** of the nitrogen gas, $V_2$ which can be obtained as:

\[ P_1 \times V^{n}_1 = P_2 \times V^{n}_2\]

\[ V_2 = \left ( \dfrac{P_1\times V^{n}_1}{P_2} \right )^ {\dfrac{1}{n}}\]

Here, $n$ is the **polytropic index** of **nitrogen** and its equal to $1.4$.

\[ V_2 = \left ( \dfrac{130kPa\times (0.07 m^3)^{1.4}}{80 kPa} \right )^ {\dfrac{1}{1.4}} \]

\[ V_2 = 0.0990 m^3 \]

Since we have obtained the **final volume,** we can calculate the **final temperature** with the formula:

\[ \dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}\]

\[ T_2 = \dfrac{V_2\times T_1}{V_1} \]

\[ T_2 = \dfrac{0.0990\times (180+273)}{0.07} \]

\[ T_2 = 640 K \]

Now we can finally calculate the **boundary** **work** **done** for the **polytropic process** using the formula:

\[ W = \dfrac{P_2 \times V_2 – P_1 \times V_1}{1-n} \]

**Substituting** the values:

\[ W = \dfrac{80k \times 0.0990 – 130k \times 0.07}{1 – 1.4} \]

\[ W = 2.95 kJ\]

Hence, the **work done**.

## Numerical Result

The **final Temperature** $T_2$ comes out to be $640 K$ whereas the **boundary work done** comes out to be $2.95 kJ$.

## Example

A **piston-cylinder** machine initially contains $0.4 m^3$ of **air** at $100 kPa$ and $80^{ \circ}C$. The air is now **isothermally condensed **to $0.1 m^3$. Find the **work done** during this process in $kJ$.

From the **statement,** we are given the following information:

The** initial Volume**, $V_1 = 0.4 m^3$.

The **initial Temperature**, $T_1 = 80^{ \circ}C = 80 + 273 = 353 K$.

The **initial Pressure**, $P_1 = 100 kPa$.

The **final Volume**, $V_2 = 0.1 m^3$.

We can calculate the **boundary work done** using the formula:

\[ W = P_1\times V_1 \log_{e}\dfrac{V_2 }{V_1}\]

\[ W = 100\times 0.4 \log_{e}\dfrac{0.1 }{0.4}\]

\[ W = -55.45 kJ \]

Note that the **negative sign** shows that the **work done** through the **system** is **negative.**