JEE PYQ: Thermodynamics Question 32
Question 32 - 2019 (10 Jan Shift 1)
A process has $\Delta H = 200$ J mol$^{-1}$ and $\Delta S = 40$ JK$^{-1}$mol$^{-1}$. Out of the values given below, choose the minimum temperature above which the process will be spontaneous:
(1) 20 K
(2) 12 K
(3) 5 K
(4) 4 K
Show Answer
Answer: (3)
Solution
For spontaneous reaction, $\Delta G < 0$. $\Delta H - T\Delta S < 0$. $\dfrac{\Delta H}{\Delta S} < T$: $\dfrac{200}{40} < T$. $5 < T$. So, minimum temperature is 5 K.
BETA
