Question

The heat capacity of chloroform (trichloromethane,CHCl3)

in the range 240K to 330K is given
byCpm/(JK-1mol-1) = 91.47
+7.5x10-2(T/K). In a particular experiment,
1.0molCHCl3 is heated from 273K to 300K. Calculate the
changein molar entropy of the sample.

Answer 1

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){(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* 10^(-2)(T/K)`

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

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

`\Delta S=1.0* [91.47\ln T+7.5* 10^(-2)T]^(300)_(273)`

`\Delta S=1.0* 91.47\ln ((T_f)/(T_i))+7.5* 10^(-2)(T_f-T_i)`

`\Delta S=1.0* 91.47\ln ((300)/(273))+7.5* 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