A) \[23.03\text{ }cal\text{ }de{{g}^{-1}}\]
B) \[-23.03\text{ }cal\text{ }de{{g}^{-1}}\]
C) \[2.303\text{ }cal\text{ }de{{g}^{-1}}\]
D) \[-2.303\text{ }cal\text{ }de{{g}^{-1}}\]
Correct Answer: B
Solution :
[b] Given, \[{{C}_{P}}=10\] cals at 1000 K \[{{T}_{1}}=1000K.\,{{T}_{2}}=100K\] \[m=32\,g\] \[\Delta S=?\] at constant pressure \[\Delta S={{C}_{p}}\] In \[\frac{{{T}_{2}}}{{{T}_{1}}}\] \[=2.303\times {{C}_{p}}log\frac{{{T}_{2}}}{{{T}_{1}}}\] \[=2.303\times 10log\frac{100}{1000}\] \[=-23.03\text{ }cal\text{ }de{{g}^{-1}}\]You need to login to perform this action.
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