Thermodynamics
Derive an expression for entropy for an ideal gas. Consider the internal energy written using its natural independent variables: dU = TdS - pdV. We also know for an ideal gas that dU = Cv dT.
a) 1. Why is the second expression true?
2. When the expressions are combined, we get Cv dT = TdS - pdV. Then we solve for S, dS = (Cv/T) dT + (p/T) dV. We can no longer integrate the expression. Explain why not.
3. Assume that Cv is constant. Integrate dS = (Cv/T) dT + (nR/V) dV.
b) Evaluate the change in entropy for an isothermal, reversible expansion in which the final volume is 50% larger than the initial volume for 1 mole. Explain.
c) Evaluate the change in entropy for an adiabatic, reversible expansion in which the final volume is 50% larger than the initial volume for 1 mole.
d) Evaluate the change in entropy for an adiabatic, irreversible expansion in which the final volume is 50% larger than the initial volume for 1 mole. Hint: the temperature is constant for an adiabatic irreversible process.