Ideal gases! Calculate the change in entropy for one mole of ideal gas which exp

Question

Ideal gases! Calculate the change in entropy for one mole of ideal gas which expands from an initial volume of 2 L and initial temperature of 500 K to a final volume of 6 L under the following conditions. P_initial refers to the pressure when T_initial = 500 K, V_initial = 2 L. Isothermal and reversible expansion. Irreversible expansion against a constant (and final) pressure of P_initial. Irreversible expansion against a vacuum...a 'free expansion' Adiabatic irreversible expansion against a constant pressure of P_final. Adiabatic reversible expansion.

Explanation / Answer

Entropy is a function of state -- the change in entropy only depends on the conditions characterizing the initial and final states of the system. For both the reversible and irreversible cases, the system starts and ends in the same state (same volume and temperature), so the entropy change for the system will be the same in both cases.

Writing the entropy as a function of volume and temperature:

S = S(V,T)

The total differential of S is then given by:

dS = (S/V)_T dV + (S/T)_V dT

For a constant-temperature process, dT = 0, so:

dS = (S/V)_T dV

This partial derivative can be expressed using variables that are usually more convenient to deal with by using a Maxwell's relation (see source):

(S/p)_T = (p/T)_V

so:

dS = (p/T)_V

For an ideal gas, p= n*R*T/V, so (p/T)_V = n*R/V, and:

dS = n*R/V dV

Integrating this gives:

S = n*R*ln(V_final/V_initial)

So the entropy change for 1 mole of ideal gas that expands from 2 liters to 6 liters at 500K is:

S = (1 mol)*(8.314 J/(mol*K))*ln(6/2) = 9.133 J/K

Now let's calculate the entropy change of the surroundings.

You should know that the internal energy of an ideal gas depends only on temperature, and because we are dealing with an isothermal process here, the internal energy of the system doesn't change (dE = 0). From the first law, we know that:

dE_sys = q_sys + w_sys

where q_sys is the thermal energy added to the system by the surroundings, and w_sys is the work done on the system by the surroundings.

and for strictly mechanical work, w_sys = - p_ext dV_sys, where p_ext is the external pressure on the system

dE_sys = q_sys - p_ext*dV_sys

Setting dE equal to zero, we get that:

q_sys = p_ext dV_sys

so

Noting that q_system = -q_surroundings, we have that:

q_surr = -p_ext dV_sys

and

dS_surr = q_surr/T_surr = -(p_ext/T_surr) dV_sys

For the reversible case, the external pressure must always be equal to the internal pressure of the system, so for this case of an ideal gas:

p_ext/T = -n*R/V_sys

dS_surr = -n*R/V_sys dV_sys

S_surr = -n*R/ln(V_sys_final/V_sys_initial) = -S_sys

So for the reversible case, the entropy change of the surroundings is equal in magnitude, but opposite in sign to the entropy change of the system, and the entropy change of the universe (the sum of the entropy changes of the system and surroundings) is zero.

For the irreversible case in this question, we are told that p_ext = Pinitial/2 atm, so:

dS_surr = -(P/2atm/500K) dV_sys

Because both p_ext and T are constant, we can easily integrate this to get:

S_surr = -(P/2atm/500K)*V_sys = -(P/2atm/500K)*(2 liter - 6 liter)

S_surr = -0.4Pinitial J/K

The entropy change of the universe is then:

S_total = S_sys + S_surr = 9.113 J/K - 0.4Pinitial J/K

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