JEE PYQ: Thermodynamics Question 31
Question 31 - 2019 (09 Jan Shift 2)
The entropy change associated with the conversion of 1 kg of ice at 273 K to water vapours at 383 K is:
(Specific heat of water liquid and water vapour are 4.2 kJ K$^{-1}$ kg$^{-1}$ and 2.0 kJ K$^{-1}$ kg$^{-1}$; heat of liquid fusion and vapourisation of water are 334 kJ kg$^{-1}$ and 2491 kJ kg$^{-1}$, respectively).
($\log 273 = 2.436$, $\log 373 = 2.572$, $\log 383 = 2.583$)
(1) 7.90 kJ kg$^{-1}$ K$^{-1}$
(2) 2.64 kJ kg$^{-1}$ K$^{-1}$
(3) 8.49 kJ kg$^{-1}$ K$^{-1}$
(4) 9.26 kJ kg$^{-1}$ K$^{-1}$
Show Answer
Answer: (4)
Solution
$\Delta S_{\text{fus}} = \dfrac{\Delta H_{\text{fus}}}{273} = \dfrac{334}{273} = 1.22$. $\Delta S_{\text{vap}} = \dfrac{\Delta H_{\text{vap}}}{373} = \dfrac{2491}{373} = 6.67$. $\Delta S_{\text{water}} = mC\ln\dfrac{T_2}{T_1} = 4.2 \times \ln\dfrac{373}{273} = 1.31$. $\Delta S_{\text{vap}} = mC\ln\dfrac{T_2}{T_1} = 2 \times \ln\dfrac{383}{373} = 0.05$. Total entropy change $\Delta S = 9.26$ kJ kg$^{-1}$ K$^{-1}$.
BETA
