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Vesnalui [34]
3 years ago
14

The heat capacity of chloroform (trichloromethane,CHCl3)

Chemistry
1 answer:
allsm [11]3 years ago
5 0

Answer : The change in molar entropy of the sample is 10.651 J/K.mol

Explanation :

To calculate the change in molar entropy we use the formula:

\Delta S=n\int\limits^{T_f}_{T_i}{\frac{C_{p,m}dT}{T}

where,

\Delta S = change in molar entropy

n = number of moles = 1.0 mol

T_f = final temperature = 300 K

T_i = initial temperature = 273 K

C_{p,m} = heat capacity of chloroform = 91.47+7.5\times 10^{-2}(T/K)

Now put all the given values in the above formula, we get:

\Delta S=1.0\int\limits^{300}_{273}{\frac{(91.47+7.5\times 10^{-2}(T/K))dT}{T}

\Delta S=1.0\times [91.47\ln T+7.5\times 10^{-2}T]^{300}_{273}

\Delta S=1.0\times 91.47\ln (\frac{T_f}{T_i})+7.5\times 10^{-2}(T_f-T_i)

\Delta S=1.0\times 91.47\ln (\frac{300}{273})+7.5\times 10^{-2}(300-273)

\Delta S=8.626+2.025

\Delta S=10.651J/K.mol

Therefore, the change in molar entropy of the sample is 10.651 J/K.mol

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