Thermodynamics and Thermochemistry - Result Question 54
54. Match the thermodynamic processes given under Column I with the expressions given under Column II.
| Column I | Column II | ||
|---|---|---|---|
| A. | Freezing of water at $273$ K and $1$ atm | p. | $q=0$ |
| B. | Expansion of $1 $ mole of an ideal gas into a vacuum under isolated conditions | q. | $W=0$ |
| C. | Mixing of equal volumes of two ideal gases at constant temprature and pressure in an isolated container | r. | $\Delta S_{sys}< 0$ |
| D. | Reversible heating of $H_2(g) $ at $1$ atm from $300$ K to $600$ K, followed by reversible cooling to $300$ K at $1$ atm. | s. | $\Delta U=0$ |
| t. | $\Delta G = 0$ |
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Answer:
Correct Answer: 54. $\mathrm{A} \rightarrow \mathrm{r}, \mathrm{t} ; \mathrm{B} \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{s} ; \mathrm{C} \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{s} ; \mathrm{D} \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{s}, \mathrm{t}$
Solution:
$\mathrm{A} \rightarrow \mathrm{r}, \mathrm{t} ; \mathrm{B} \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{s} ; \mathrm{C} \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{s} ; \mathrm{D} \rightarrow \mathrm{p}, \mathrm{q}, \mathrm{s}, \mathrm{t}$
(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)
$\Delta U=0$
$\because T _i=T _f, \Delta G=0$
BETA
