Thermodynamics and Thermochemistry - Result Question 54
####54. Match the thermodynamic processes given under Column I with the expressions given under Column II.
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Answer:
Correct Answer: 54. $A \rightarrow p, r, s ; B \rightarrow r, s ; C \rightarrow t ; D \rightarrow p, q, t$
Solution:
- (A) $\rightarrow r, t$; (B) $\rightarrow p, q, s$; (C) $\rightarrow p, q, s ;(D) \rightarrow q, s, t$
(A) $H _2 O(l) \underset{1 \text { atm }}{\stackrel{0^{\circ} C}{\rightleftharpoons}} H _2 O(s)$
$q<0, W<0$ (expansion)
$\Delta S _{\text {sys }}<0$ (solid state is more ordered than liquid state)
$\Delta U<0 ; \Delta G=0$ (At equilibrium)
(B) $q=0$ (isolated), $W=0\left(p _{\text {ext }}=0\right)$
$\Delta S _{\text {sys }}>0 \quad \because V _2>V _1$
$\Delta U=0 \because q=W=0$
$\Delta G<0 \because p _2<p _1$
(C) $q=0$ (isothermal mixing of ideal gases at constant $p$ )
$W=0 \because \Delta U=0 ; q=0, \Delta S _{\text {sys }}>0$
$\because V _2>V _1, \Delta U=0$
$\because \Delta T=0$
$\Delta G<0 \because$ mixing is spontaneous.
(D) $q=0$ (returning to same state and by same path)
$$ W=0 $$
$\Delta S _{\text {sys }}=0$ (same initial and final states)
$\Delta U=0$
$\because T _i=T _f, \Delta G=0$
