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The ordinate and abscissa of the diagram to depict the isobaric processes of an ideal gas as a hyperbola are, respectively

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ESE Mechanical 2018 Official Paper
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  1. temperature and entropy
  2. Internal energy and volume
  3. temperature and density
  4. enthalpy and entropy

Answer (Detailed Solution Below)

Option 3 : temperature and density
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Detailed Solution

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Concept:

The ideal gas equation is given by

PV = mRT

\(\Rightarrow P = \frac{m}{V}RT = \rho RT\)

P = ρRT 

For isobaric process

P1 = P2

∴ ρT = Constant, which is the equation for rectangular hyperbola. 

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