- ✓The enthalpy of mixing is zero
- BThe entropy of mixing is zero
- CThe free energy of mixing is zero
- DThe free energy as well as the entropy of mixing are each zero
$a) V _{\text {mix }}=0$
$b) H _{\text {mix }}=0$
$c) \Delta G _{\text {mix }}=-v e$
Final Answer : Hence, option $A$ is correct.
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$1 \mathrm{~mol}$ of an ideal gas is kept in a cylinder, fitted with a piston, at the position $\mathrm{A}$, at $18^{\circ} \mathrm{C}$. If the piston is moved to position $B$, keeping the temperature unchanged, then ' $x$ ' $L$ atm work is done in this reversible process.
$\mathrm{x}=$ . . . . . . $\mathrm{L} \mathrm{atm.} \mathrm{(nearest} \mathrm{integer)}$
[Given : Absolute temperature $={ }^{\circ} \mathrm{C}+273.15$, $\left.\mathrm{R}=0.08206 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right]$
$Z{n^{2 + }} + 2{e^ - } \to Zn$ ; $E = - 7.62\,\,V,$
$F{e^{2 + }} + 2{e^ - } \to Fe$ ; $E = - 7.81\,\,V$
The emf of the cell $F{e^{2 + }} + Zn \to Z{n^{2 + }} + Fe$ is ............ $\mathrm{V}$
What is the value of $x$ in the above reaction ?