Question

A piston-cylinder device initially contains $0.07{{m}^{3}}$of nitrogen gas at 130 kPa and $120{}^\circ C$. The nitrogen is now expanded polytropically to a state of 100 kPa and $100{}^\circ C$. Determine the Boundary work of the process.
A. 1.86 kJ
B. 1.68 kJ
C. 6.81 kJ
D. 8.18 kJ

Answer 1

Using Boyle’s law we will find the initial mass of the nitrogen gas. Using ideal gas law we will find the initial volume of the nitrogen gas. Using Boyle’s law we will determine the value of the polytropic index for the nitrogen. Finally, we will compute the work done using polytropic processes equations.

Formula used: ${{p}_{1}}.{{V}_{1}}^{n}={{p}_{2}}.{{V}_{2}}^{n}$
$W=\dfrac{{{p}_{2}}.{{V}_{2}}-{{p}_{1}}.{{V}_{1}}}{1-n}$
$m=\dfrac{{{p}_{1}}.{{V}_{1}}}{R.{{T}_{1}}}$

Complete step by step answer:
From given, we have
Firstly, compute the value of the final volume.
Consider the data as follows:
The initial volume is, ${{V}_{1}}=0.07{{m}^{3}}$
The initial pressure is, ${{p}_{1}}=130\,kPa$
The final pressure is, ${{p}_{2}}=80\,kPa$
The mass and the volume of the nitrogen at the initial state are calculated as follows.
Using the ideal gas law, we will determine the mass of the nitrogen gas.
So, we get,

& m=\dfrac{{{p}_{1}}.{{V}_{1}}}{R.{{T}_{1}}} \\\ & \Rightarrow m=\dfrac{130k\times 0.07}{0.2986\times (120+273)} \\\ & \Rightarrow m=0.07802\,kg \\\ \end{aligned}$$ Therefore, the initial mass of the nitrogen gas is obtained to be equal to 0.07802 kg. The initial volume of the nitrogen gas is obtained again by using the ideal gas law. So, we get, $$\begin{aligned} & {{V}_{2}}=\dfrac{mR{{T}_{2}}}{{{p}_{2}}} \\\ & \Rightarrow {{V}_{2}}=\dfrac{0.07802\times 0.2968\times (100+273)}{100\,k} \\\ & \Rightarrow {{V}_{2}}=0.08637\,{{m}^{3}} \\\ \end{aligned}$$ Therefore, the initial volume of the nitrogen gas is obtained to be equal to $$0.08637\,{{m}^{3}}$$ Using Boyle’s law equation, we will determine the value of the polytropic index for the nitrogen. So, we get, $${{p}_{1}}.{{V}_{1}}^{n}={{p}_{2}}.{{V}_{2}}^{n}$$ Substitute the obtained values in the above equation. $$\begin{aligned} & 130\,k\times {{(0.07)}^{n}}=100\,k\times {{(0.08637)}^{n}} \\\ & \Rightarrow n=1.249 \\\ \end{aligned}$$ The boundary work done is calculated using the formula for polytropic processes $$W=\dfrac{{{p}_{2}}.{{V}_{2}}-{{p}_{1}}.{{V}_{1}}}{1-n}$$ Substitute the values in the above equation. $$\begin{aligned} & W=\dfrac{100\,k\times 0.08637-130\,k\times 0.07}{1-1.249} \\\ & \Rightarrow W=1.86\,kJ \\\ \end{aligned}$$ Hence the work done. **So, the correct answer is “Option A”.** **Note:** The things to be on your finger-tips for further information on solving these types of problems are: The units of the given parameters should be taken into consideration while solving the problem. In this case, the polytropic index is considered to be the same as that of the number of moles of a gas.