Answer
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Hint: A specific portion of matter with a definite boundary on which is observed is a thermodynamic system. A closed surface surrounding a system through which energy and mass enter or leave the system is a boundary and interacts with the system's surroundings. The capacity to do by the system or on the system in a thermodynamic system is said to be work done.
Complete step by step answer:
- There are many thermodynamic processes in the system that have unique properties. The given system is an isothermal reversible expansion process, which represents work under the area of PV-curve.
- For calculating work done, calculate the area under the PV – curve for solid and dotted lines.
Given, ${{W}_{s}}$ is the work done along the solid line path, and ${{W}_{d}}$ is the work done along the dotted lines in the curve.
From the graph, ${{W}_{d}}=4\times 1.5+1\times 1+2.5\times \dfrac{2}{3}=8.65$
Hence, the process is isothermal reversible expansion,
Then ${{W}_{s}}=-2.303nRT\log \dfrac{{{v}_{2}}}{{{v}_{1}}}$
From the graph, ${{W}_{s}}=2\times 2.303\log \dfrac{5.5}{0.2}=4.79$
Then the integer close to the ratio is $\dfrac{{{W}_{d}}}{{{W}_{s}}}=\dfrac{8.65}{4.79}=1.80\simeq 2$
So, the correct answer is “Option B”.
Note: Isothermal process means the temperature of the working substance during expansion or compression remains constant and during this process, there is no change in the temperature and no change in the internal energy. In the isothermal process, the external pressure constant is irreversible expansion, while the external pressure decreases continuously infinitesimally smaller than internal pressure.
Complete step by step answer:
- There are many thermodynamic processes in the system that have unique properties. The given system is an isothermal reversible expansion process, which represents work under the area of PV-curve.
- For calculating work done, calculate the area under the PV – curve for solid and dotted lines.
Given, ${{W}_{s}}$ is the work done along the solid line path, and ${{W}_{d}}$ is the work done along the dotted lines in the curve.
From the graph, ${{W}_{d}}=4\times 1.5+1\times 1+2.5\times \dfrac{2}{3}=8.65$
Hence, the process is isothermal reversible expansion,
Then ${{W}_{s}}=-2.303nRT\log \dfrac{{{v}_{2}}}{{{v}_{1}}}$
From the graph, ${{W}_{s}}=2\times 2.303\log \dfrac{5.5}{0.2}=4.79$
Then the integer close to the ratio is $\dfrac{{{W}_{d}}}{{{W}_{s}}}=\dfrac{8.65}{4.79}=1.80\simeq 2$
So, the correct answer is “Option B”.
Note: Isothermal process means the temperature of the working substance during expansion or compression remains constant and during this process, there is no change in the temperature and no change in the internal energy. In the isothermal process, the external pressure constant is irreversible expansion, while the external pressure decreases continuously infinitesimally smaller than internal pressure.
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