At Constant volume By first law of thermodynamics, \(\triangle U=q+W \ or \ q=\triangle U-W.\)But  \(W=-p\triangle V.\) Hence, \(q=\triangle U+p\triangle V.\) But as volume remains constant, \(\triangle V=0\) .
\(\therefore qV=\triangle U\). But \(\triangle U\) is state function. Hence, \(q_{ v }\) is state function.

18.0 g H₂O = 1 mol H₂O
Enthalpy change for vaporising 1 mole of H₂O = 40.79 KJ
\(\therefore \) Enthalpy change for vaporising 2 moles of 
H₂O = 2 x 40.79 KJ = 81.58 KJ
Standard enthalpy of vaporisation at 100^oC and 1 bar pressure,
\({ \triangle }_{ vap }{ H }^{ \circ }=+40.79 \ KJ \ { mol }^{ -1 }\)

\(Work \ done,\ W=-{ p }_{ ext }({ V }_{ 2 }-{ V }_{ 1 })\)
\(As \ { p }_{ ext }=0,\ so \ w=-0(5-1)=0\)
\( For \ isothermal \ expansion,\)
\( \triangle U=0 \ as \ \triangle T=0\)

In the first case, as the expansion is against constant external pressure,
\(w=-{ p }_{ ext }({ V }_{ 2 }-{ V }_{ 1 })\)
\( =-2bar\times (50-10)L\)
\( =-80Lbar=-80\times 100J\)
\( =-8kJ\)
If the above expansion were carried out reversibly, the internal pressure of the gas should be infinitesimally greater that external pressure at every stage. Hence, the work done will be more.

1 mole of CCI₄ = 154 g
Heat required for vapourising 154 g CCI₄ = 30.5 kJ
\(\therefore\) Heat required for vapourising
\(284 \ g \ CC{ I }_{ 4 }=\frac { 30.5\times 284 }{ 154 } kJ=56.25kJ\)