Match the following:
$A$. Isothermal process$i$. $q = \Delta U$
$B$. Adiabatic process$ii$. $W = - P \times \Delta V$
$C$. Isobaric process$iii$. $W = \Delta U$
$D$. Isochoric process$iv$. $W = - nRT \ln \left(\frac{v_f}{v_i}\right)$

  • A
    $A-iv, B-iii, C-ii, D-i$
  • B
    $A-iii, B-iv, C-i, D-ii$
  • C
    $A-i, B-ii, C-iii, D-iv$
  • D
    $A-ii, B-i, C-iv, D-iii$

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The entropy change associated with the conversion of $1 \ kg$ of ice at $273 \ K$ to water vapours at $383 \ K$ is: (Specific heat of water liquid and water vapours are $4.2 \ kJ \ K^{-1} \ kg^{-1}$ and $2.0 \ kJ \ K^{-1} \ kg^{-1},$ heat of fusion and vaporisation of water are $334 \ kJ \ kg^{-1}$ and $2491 \ kJ \ kg^{-1},$ respectively) $(\ln \ 273 = 5.61, \ln \ 373 = 5.92, \ln \ 383 = 5.95)$

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$AB \to$ Isothermal expansion
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Molar enthalpy change for vapourisation of $1.0 \ mol$ of water at $1.0 \ bar$ and $100 ^{\circ} C$ is $41.0 \ kJ \ mol^{-1}$. If water vapour is assumed to be an ideal gas,the internal energy change for $1.0 \ g$ of water in $kJ$ is

For an endothermic reaction where $\Delta H$ represents the enthalpy of the reaction in $kJ/mole$,the minimum value for the energy of activation $(E_a)$ will be:

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