1.

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):}`

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.


Discussion

No Comment Found

Related InterviewSolutions