Calculate the change in entropy for converting 1 mole gaseous benzene to liquid benzene at
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Information has been given that: The boiling point of benzene at atmospheric pressure is 353 K353 K and the enthalpy of vaporization of benzene is 30.8 kJ mol−130.8 kJ mol−1 at this temperature. The molar heat capacities of the liquid and vapour are 136.1 J K−1 mol−1136.1 J K−1 mol−1 and 81.7 J K−1 mol−181.7 J K−1 mol−1, respectively, and may be assumed temperature independent.
My question is how can I work out the entropy change of the system, the surroundings and hence the universe when 1 mol1 mol of benzene vapour at 343 K343 K and atmospheric pressure becomes liquid benzene at 343 K343 K. Also, will this process occur spontaneously?
My working so far:
I know that dS=dHTdS=dHT therefore,
dS=−30.8×103 J343 K=−89.8 J K−1dS=−30.8×103 J343 K=−89.8 J K−1 which is the entropy of the system.
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My question is how can I work out the entropy change of the system, the surroundings and hence the universe when 1 mol1 mol of benzene vapour at 343 K343 K and atmospheric pressure becomes liquid benzene at 343 K343 K. Also, will this process occur spontaneously?
My working so far:
I know that dS=dHTdS=dHT therefore,
dS=−30.8×103 J343 K=−89.8 J K−1dS=−30.8×103 J343 K=−89.8 J K−1 which is the entropy of the system.
Hope this question's answer are helpful
And press the point of thanks
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