A piston–cylinder device initially contains 0.07m^{3} of nitrogen gas at 1

Given:
Initial volume ($V_1$) = 0.07 m${}^3$
Initial pressure ($P_1$) = 130 kPa
Initial temperature ($T_1$) = 180°C
Final pressure ($P_2$) = 80 kPa
Polytropic exponent ($n$) = specific heat ratio ($\gamma$)
To find:
Final temperature ($T_2$) and boundary work done ($W_{\text{boundary}}$)
The process is an isentropic expansion, which means the entropy remains constant ($\Delta S=0$).
Using the ideal gas law, we can write:
$PV=mRT$
Where:
$P$ is the pressure
$V$ is the volume
$m$ is the mass of the gas
$R$ is the specific gas constant
$T$ is the temperature
Assuming the mass of the nitrogen gas remains constant, we can rewrite the equation as:
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
Since the process is isentropic, we also have:
$\frac{P_1}{P_2}={\left(\frac{V_2}{V_1}\right)}^{\gamma }$
Solving these two equations simultaneously, we can find $T_2$ and $V_2$.
Now, let's solve for $V_2$:
$\frac{P_1}{P_2}={\left(\frac{V_2}{V_1}\right)}^{\gamma }$
Taking the logarithm of both sides:
$\log \left(\frac{P_1}{P_2}\right)=\gamma \log \left(\frac{V_2}{V_1}\right)$
Solving for $V_2$:
$\frac{V_2}{V_1}={\left(\frac{P_1}{P_2}\right)}^{1/\gamma }$
Substituting the given values:
$\frac{V_2}{0.07}={\left(\frac{130}{80}\right)}^{1/\gamma }$
Simplifying further, we have:
$V_2=0.07\times {\left(\frac{130}{80}\right)}^{1/\gamma }$
Now, let's solve for $T_2$:
$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$
Solving for $T_2$:
$T_2=\frac{P_2{V}_2{T}_1}{P_1{V}_1}$
Substituting the given values:
$T_2=\frac{80\times 0.07\times 180}{130\times 0.07}=\frac{80\times 180}{130}$
Simplifying further, we have:
$T_2=\frac{14400}{130}$
Finally, let's calculate the boundary work done ($W_{\text{boundary}}$):
$W_{\text{boundary}}=\frac{P_2{V}_2-P_1{V}_1}{1-n}$
Substituting the given values:
$W_{\text{boundary}}=\frac{80\times 0.07-130\times 0.07}{1-\gamma }$
Simplifying further, we have:
$W_{\text{boundary}}=0.07\times \frac{80-130}{1-\gamma }$
Therefore, the final temperature $T_2$ is $\frac{14400}{130}$ and the boundary work done $W_{\text{boundary}}$ is $0.07\times \frac{80-130}{1-\gamma }$.

Step 1: Calculate the initial temperature using the ideal gas equation.
Given:
$V_1=0.07{\text{m}}^3$ (initial volume)
$P_1=130\text{kPa}$ (initial pressure)
$T_1={180}^{\circ }\text{C}=180+273.15\text{K}$ (initial temperature)
We can rearrange the ideal gas equation to solve for $T$:
$T=\frac{P·V}{m·R}$
The mass $m$ is unknown, but we can use the ideal gas equation to find it:
$PV=mRT$
Solving for $m$ gives:
$m=\frac{PV}{RT}$
Substituting this value of $m$ back into the equation for temperature, we have:
$T=\frac{P·V}{\left(\frac{PV}{RT}\right)·R}=\frac{T}{P}$
Plugging in the given values, we get:
$T_1=\frac{180+273.15}{130}\text{K}$
Step 2: Calculate the final temperature using the polytropic process equation.
Given:
$P_2=80\text{kPa}$ (final pressure)
$n=\gamma$ (polytropic exponent, where $\gamma$ is the specific heat ratio)
Using the polytropic process equation, we can write:
$P_1·V_1^n=P_2·V_2^n$
Since the initial and final volumes are not given, we can write:
$\frac{V_2}{V_1}={\left(\frac{P_1}{P_2}\right)}^{1/n}$
$V_2=V_1·{\left(\frac{P_1}{P_2}\right)}^{1/n}$
Now, let's calculate the final volume $V_2$ using the given values:
$V_2=0.07{\text{m}}^3·{\left(\frac{130}{80}\right)}^{1/\gamma }$
Step 3: Calculate the final temperature using the ideal gas equation.
Now that we have the final volume $V_2$, we can use the ideal gas equation to calculate the final temperature $T_2$:
$T_2=\frac{P_2·V_2}{m·R}$
Step 4: Calculate the boundary work done during the process.
The boundary work done during the process can be calculated using the formula:
$W={\int }_{V_1}^{V_2}PdV$
Since the process is isentropic, we can use the equation:
$P{V}^n=\text{constant}$
To write:
$P=\frac{\text{constant}}{V^n}$
Substituting this into the work equation, we have:
$W={\int }_{V_1}^{V_2}\frac{\text{constant}}{V^n}dV$
Simplifying, we get:
$W=\frac{\text{constant}}{n-1}(V_2^{1-n}-V_1^{1-n})$
Since we already have the values of $V_1$ and $V_2$, we can calculate the work using the given polytropic exponent $n=\gamma$.