$$\eqalign{ & {\rm{Solid}}\buildrel {{\rm{Sublimation}}} \over \longrightarrow {\rm{Vapour}} \cr & {\rm{This}}\,\,{\rm{process}}\,\,{\rm{occurs}}\,\,{\rm{in}}\,\,{\rm{two}}\,\,{\rm{steps,}} \cr & {\rm{Step 1:}}\,\,{\rm{Solid}}\buildrel {{\rm{Fusion}}} \over \longrightarrow {\rm{liquid}} \cr & {\rm{Step 2: Liquid}}\buildrel {{\rm{Vaporisation}}} \over \longrightarrow {\rm{Vapour}} \cr} $$
Thus, Enthalpy of sublimation = Enthalpy of fusion + Enthalpy of vaporisation.

We know that $$\Delta G = \Delta H - T\Delta S$$
When $$\Delta H < 0$$   and $$\Delta S < 0$$   then $$\Delta G$$  will be negative at low temperatures ( positive at high temperature ) and the reaction will be spontaneous.

$$\eqalign{ & {\text{Given,}}\,\Delta H = 170\,kJ = 170 \times {10^3}J \cr & \Delta S = 170\,J{K^{ - 1}};\,T = ? \cr & \Delta G = \Delta H - T\Delta S \cr & {\text{For spontaneous reaction,}} \cr} $$
$$\Delta G < 0 \Rightarrow 0 < 170 \times {10^3} - T \times 170\,;$$        $$T > 1000$$
$$\therefore \,\,T = 1110\,K$$

This is in accordance with Hess's law.

$${q_p} = n{C_p}\Delta T;{q_p} = 10\,kJ\,\,{\text{or}}\,\,10000\,J$$
$${C_p} = 75\,J\,{K^{ - 1}}\,mo{l^{ - 1}}.$$     $${\text{No}}{\text{. of moles}},\,n = \frac{{1000}}{{18}} = 55.5\,mol$$
$$\Delta T = \frac{{10000}}{{55.5 \times 75}} = 2.4\,K$$

$$q = - w = {P_{ex}}\left( {10 - 2} \right) = 0\left( 8 \right) = 0$$
For isothermal expansion in vacuum, $${P_{ex}} = 0$$

The specific heat is an intensive property which does not depend on the quantity or size of matter.

Work done,
$$\eqalign{ & w = - {P_{ext.}}\,\,dV \cr & w = - 2 \times 2.5 \cr & \,\,\,\,\,\, = - \,5\,L\,atm \cr & \,\,\,\,\,\, = - 506.3\,J \cr} $$
Because this work is used in raising the temperature of water, so work done is equal to the heat supplied i.e.,
$$w = q = m \cdot {c_s} \cdot \Delta T$$
Given that, $$m = 18\,g\left( { = 1\,mole} \right),{c_s} = 4.184\,J\,{g^{ - 1}}\,{K^{ - 1}},$$         $$q = + 506.3\,J\,$$   ( Heat is given to water ), $$\Delta T = ?$$
$$\eqalign{ & \Delta T = \frac{q}{{{c_s} \cdot m}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{506.3}}{{4.184 \times 18}} \cr & \,\,\,\,\,\,\,\,\, = 6.72 \cr} $$
∴ Final temperature,
$$\eqalign{ & {T_f} = {T_i} + \Delta T \cr & \,\,\,\,\,\,\,\, = 293 + 6.72 \cr & \,\,\,\,\,\,\,\, = 299.72\,K \approx 300\,K \cr} $$

$$\eqalign{ & \Delta H = \sum {B.E{._R}} - \sum {B.E{._P}} \cr & \,\,\,\,\,\,\,\,\,\,\, = \left( {2 \times 44 + 2 \times 111} \right) - \left( {119 + 2 \times 135} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = - 79\,kcal \cr} $$

$$\Delta H = n{C_p}\,\Delta T$$    solution; since $$\Delta T = 0\,{\text{so}},\,\,\Delta H = 0$$