Derive the entropy of an ideal gas

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SUMMARY

The entropy of an ideal gas can be derived using the equation ΔS = ∫(S₀ to S) dS = ∫(T₀ to T) (∂S/∂V) dT + ∫(V₀ to V) (∂S/∂V) dV, under the condition that the molar specific heat at constant volume (Cₕ) remains constant. By integrating from 0 to T, the relationship dQ = TdS = dU + PdV simplifies to nCₕdT = TdS, leading to the equation ∫₀^T dS = ∫₀^T nCₕ dT/T. Assuming the entropy at 0 K is zero (S₀ = 0), the resulting expression for ST represents the entropy of the gas at temperature T.

PREREQUISITES
  • Understanding of thermodynamics principles
  • Familiarity with the concept of molar specific heat
  • Knowledge of integration techniques in calculus
  • Basic concepts of entropy in statistical mechanics
NEXT STEPS
  • Study the derivation of the first and second laws of thermodynamics
  • Learn about the implications of constant specific heat in thermodynamic processes
  • Explore advanced integration techniques relevant to thermodynamic equations
  • Investigate the relationship between entropy and temperature in ideal gases
USEFUL FOR

Students and professionals in physics and engineering, particularly those specializing in thermodynamics and statistical mechanics, will benefit from this discussion.

Lee
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THe Question asks 'Derive the entropy of an ideal gas when its molar specific heat at constant volume is constant.'

So I've taken

\Delta S = \int_{S_0}^{S} dS = \int_{T_0}^{T} \frac{\partial_S} {\partial_V} dT + \int_{V_0}^{V} \frac{\partial_S}{\partial_V} dV

in this context what would be the next best step?
 
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Lee said:
THe Question asks 'Derive the entropy of an ideal gas when its molar specific heat at constant volume is constant.'

So I've taken

\Delta S = \int_{S_0}^{S} dS = \int_{T_0}^{T} \frac{\partial_S} {\partial_V} dT + \int_{V_0}^{V} \frac{\partial_S}{\partial_V} dV

in this context what would be the next best step?
If the specific heat remains constant at all temperatures, then it is possible to integrate from temperature 0 to T.

Since dQ = TdS = dU + PdV = nC_vdT + PdV at constant volume nC_vdT = TdS

so:

\int_0^T dS = \int_0^T nC_v dT/T = S_T - S_0

If you let the entropy of the gas at 0 K be 0: S_0 = 0, then ST represents the entropy of the gas at temperature T.

AM
 

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