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Match the entries listed in Column I with appropriate entries listed in Column II. `{:(,"Column I",,"Column II"),((A),"Isothermal process",(p),((delU)/(delV))_(T)=0),((B),-nFoverset(Theta)E,(q),W=-DeltaU),((C),"Adiabatic reaction",(r),DeltaU=0),((D),"van der waals gas",(s),DeltaG^(Theta)),((E),"Ideal gas",(t),((delT)/(delP))_(H)ne0):}` |
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Answer» Correct Answer - A `rarr` t; B `rarr` r; C `rarr` p; D `rarr` p (A) For ideal gas, `((delU)/(delV))_(T)=0` (B) In adiabatic process, q = 0. Therefore, from the first law of thermodynamic, `DeltaU=q-w and q=0` `therefore DeltaU=-w` (C ) In isothermal process, `DeltaU=0`, because internal energy is the function of temperature (D) `DeltaG^(Theta)=-nFE^(Theta)` (E) `((delT)/(delP))_(H)ne0` for real gases. |
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