A) \[0.336,\text{ }1\]
B) \[0.6258,\text{ }2\]
C) \[5.528,\text{ }3\]
D) \[0.647,\text{ }0\]
Correct Answer: B
Solution :
[b] Step 1. (Using the third law of thermodynamics): (For changing \[{{H}_{2}}O(s)\,\,(-10{}^\circ C,\,1atm)\xrightarrow{{}}{{H}_{2}}O\,(s,\,0{}^\circ C\,1atm)\] \[\Delta {{S}_{1}}=\int\limits_{-10}^{0}{n\frac{{{C}_{P}}}{T}}dT=1\times 9\times 2.3\times \log \frac{273}{263}\] \[=0.336\,cal\,\,{{\deg }^{-1}}mo{{l}^{-1}}\] Step 2 (Using the second law of thermodynamics): \[{{H}_{2}}O(s)\,\,(0{}^\circ C,\,1atm)\xrightarrow{{}}{{H}_{2}}O(l)\,(0{}^\circ C,\,1atm)\] \[\Delta {{S}_{2}}=\frac{{{q}_{rev}}}{T}=\frac{1440}{273}=5.258\,cal\,\,{{\deg }^{-1}}\,mo{{l}^{-1}}\] Step 3 (Using the third law of thermodynamics): \[{{H}_{2}}O(l)\,(0{}^\circ C,\,1atm)\xrightarrow{{}}{{H}_{2}}O(l)\,\,(10{}^\circ C,\,1atm)\] \[\Delta {{S}_{3}}=\int\limits_{0}^{-10}{n\frac{{{C}_{P}}}{T}}dT=1\times 18\times 2.3\times \log \frac{283}{273}\] \[=0.647\,\,cal\,\,{{\deg }^{-1}}mo{{l}^{-1}}\] \[\Delta S=\Delta {{S}_{1}}+\Delta {{S}_{2}}+\Delta {{S}_{3}}=0.336+5.258+0.647\] \[=6.258\,cal\,\,{{\deg }^{-1}}\,mo{{l}^{-1}}\]
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