Question

Match the following

$\begin{array}{cccc}i)& \text{Entropy of vaporisation}& a)& \text{decreases}\\ ii)& \text{K for spontaneous process}& b)& \text{is alwasy positive}\\ iii)& \text{Crystalline solid state}& c)& \text{lowest entropy}\\ iv)& \Delta U\text{in adiabatic expansion of ideal gas}& d)& \frac{\Delta H_{vap}}{T_b}\end{array}$

Answer 1

i) Entropy of vaporisation is always positive as liquid changes to more random gaseous state, which have more disorder

$\Delta S=\frac{Q}{T}$ $\therefore \Delta S_{vap}=\frac{\Delta H_{vap}}{T_b}$

Where $T_b=$ Boiling point of liquid

ii) $K$ for spontaneous reaction is always positive

$\Delta G^o=-RTlnK$ ($\Delta G^o<0$ for spontaneous reaction). So, $K$ should be $>1$ for $\Delta G^o<0$

iii) Crystalline solid state has lowest entropy because of a less disordered state. As all atom/ions are located at their fixed positions in crystal lattice.

iv) $\Delta U$ adiabatic expansion of ideal gas always decreases.
As per First law of thermodynamics,

$\Delta U=w\to$ for Adiabatic process $(q=0)$

For expansion, we know work is done by system. So, work $\left(w\right)$ will be negative and according to previous equation

$\Delta U=w$

it will give negative value of $\Delta U$, which shows that internal energy will decrease during adiabatic expansion of ideal gas.

So, on basis of these explanation, the correct match of following will be

$\begin{array}{cccc}i)& \text{Entropy of vaporisation}& b)& \text{is always positive}\\ & & d)& \frac{\Delta H_{vap}}{T_b}\\ ii)& \text{K for spontaneous process}& b)& \text{is alwasy positive}\\ iii)& \text{Crystalline solid state}& c)& \text{lowest entropy}\\ iv)& \Delta U\text{in adiabatic expansion of ideal gas}& a)& \text{decreases}\end{array}$