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Show that entropy as a state function. |
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Answer» Solution :Consider a cylinder fitted with a frictionless and weightless piston, which contains a gas and is in contact with a large heat reservoir. During isothermal and REVERSIBLE expansion of the gas from volume `v_(1)` to `v_(2)`, let there be absorption of heat q at temperature T. `:.` Change in ENTROPY of the system. `Delta S_(sys) = (q_(rev))/(T)` Since an EQUIVALENT amount of hest will be lost by reservoir, change in entropy of reservoir will be `Delta S_(res) = Delta S_(sys) + Delta S_(res) = (q_(rev))/(T) + ((-q_(rev))/(T)) = 0` If we now, compress the gas isothermally from a volume `v_(2)` to `v_(1)`, heat given by the system is `-q_(rev)`. `:. Delta S_(sys) = (-q_(rev))/(T)` and `Delta S_(rev) = (q_(rev))/(T)` Total change in entropy `Delta S_(2) = (-q_(rev))/(T) + (q_(rev))/(T) = 0` Total change in entropy for the complete cycle `Delta S_(1) + Delta S_(2) = 0` At the END of the cycle, the entropy of the system is the same as it initially. Therefore, entropy is a state function. |
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