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A sample of an ideal gas is isothermally expanded at 45.0 °C and the gas does 575 J of work during this expansion. Calculate

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Answer #1

Here the temperature is constant, and we know that the internal energy is Function of temperature hence the change in internal energy will be zero.(dU=0)

From 1st law of thermodynamics,

dQ= dU + PdV

dQ= work done.

And entropy is defined as dQ/T= S, T=45 °C=318.15K

So S= 575/318.15

S= 1.8073 J/K is the answer.

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