Lacey VonRueden PhD
Oct 14, 2024
Consider a process in which an ideal gas is compressed to one-sixth of its original volume at constant temp?
Consider a process in which an ideal gas is compressed to one-sixth of its original volume while maintaining a constant temperature. What is the entropy change per mole of the gas during this process? Please provide the necessary calculations and explanations to support your answer.
1 Answers
If we define the entropy as a function of T and V, then the total differential of the entropy is given by:
dS = (∂S/∂V)_T dV + (∂S/∂T)_V dT
For an isothermal process, dT = 0, so:
dS = (∂S/∂V)_T dV
Using a Maxwell’s relation (see source) we can write (∂S/∂V)_T = (∂p/∂T)_V, so:
dS = (∂p/∂T)_V dV
For an ideal gas, p = n*R*T/V, so (∂p/∂T)_V = n*R/V, and:
dS = (n*R) * dV/V
Integrating this, we get:
ΔS = n*R*ln(V_final/V_initial)
This is the equation that defines the entropy change for an isothermal expansion of an ideal gas.
In this case, we have that n = 1 mol, and V_final/V_initial = 1/6, so:
ΔS = (1mol)*(8.314 J/(mol*K))*ln(1/6)
ΔS/mol = -14.9 J/(mol*K)
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