Question
Download Solution PDFA non-flow reversible process takes place according to , where P is in bar. What will be the work done if the pressure changes from 1 bar to 10 bar? [Given, ]
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
The process is isothermal: \( PV = \text{constant} = 15 \)
Work done in isothermal process:
\( W = \int_{P_1}^{P_2} \frac{15}{P} dP = 15 \cdot \ln\left(\frac{P_2}{P_1}\right) \)
Calculation:
\( W = 15 \cdot \ln\left(\frac{10}{1}\right) = 15 \cdot 2.3025 = 34.5375 \, \text{bar·m}^3 \)
\( = 34.5375 \times 10^5 = 3.45375 \times 10^6 \, \text{N·m} = 3.453 \, \text{MN·m} \)
Since pressure increases (volume decreases), it's compression.
Last updated on Jul 15, 2025
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