In answering this set of questions, you are encouraged to draw a PV diagram, wit

Question

In answering this set of questions, you are encouraged to draw a PV diagram, with P (pressure) on the y axis and V (volume) on the x axis. Plot the three points A, B, and C on this diagram.

Consider 7.00 liters of an ideal (monatomic) gas at a pressure of 40.0 atm and a temperature of 428 K. Call this state of the system A. Using the ideal gas law, calculate the number of moles of gas present in the system.

Number of moles, n = moles

The gas expands isothermally and reversibly to a final pressure of 16.0 atm. Call this state of the system B. Calculate the volume of the gas at state B.

Volume at state B = liters

Calculate the entropy change in the system when the system moves down the isotherm from state A to state B.

Entropy change down the isotherm = JK-1

Now consider an adiabatic and reversible expansion from state A to a state C where the volume is the same as at state B. Calculate the temperature at state C.

Temperature at state C = K

What is the entropy change in the system when the system moves down the adiabat from state A to state C?

Entropy change down the adiabat = JK-1

What is the entropy change in the system when the system moves from state B at the end of the isotherm to state C at the end of the adiabat at constant volume?

Entropy change from state B to state C = JK-1

Explanation / Answer

The ideal gas law is:

p*V = n*R*T

n = p*V/(R*T)

so 7 liters of an ideal gas at p = 40 atm and T = 428 K corresponds to:

n = (40 atm * 7 liter)/(0.082057 (liter*atm)/(K*mol) * 428K)

n = 7.9725 mol

If this is a closed system (constant number of moles), and the gas expands isothermally to attain a pressure of 10 atm (at this point, the fact that the expansion is reversible is irrelevant), the volume would be:

V = n*R*T/p = (7.9725mol)*(0.082057 (liter*atm))/(K*mol))*(428K)/(16 atm)

V = 17.5 liters

The entropy change for a reversible isothermal expansion of an ideal gas is given by:

deltaS = n*R*ln(V_final/V_initial) = n*R*ln(p_initial/p_final)

Plugging in the appropriate numbers for this case:

deltaS = (7.9725 mol)*(0.082057 (liter*atm))*ln(40/16)

deltaS = 60.0 J/K in going from A to B

For an adiabatic expansion of an ideal gas:

p*V^gamma = constant. Alternatively, because p = n*R*T/V,

T*V^(gamma - 1) = constant

where gamma = Cp/Cv, and for a monatomic ideal gas gamma = 5/3.

Initially (at state A):

constant = (428K)*(7 liters)^(5/3)

So at state C:

constant = (428K)*(7 liters)^(2/3) = (T_C)*(28 liters)^(2/3)

where T_C is the temperature at state C.

T_C = 169.85 K

For a reversible process, dS = deltaq/T and for an adiabatic process, deltaq = 0, so for a reversible, adiabatic process is an isentropic process in which dS = deltaS = 0. The entropy change following an adiabatic path from state A to state C is zero.

Entropy is a function of state, so the entropy change between two states does not depend on the path taken between those states.

If the entropy change between states A and C is zero,then:

0 = deltaS(A->C) = deltaS(A->B) + deltaS(B->C)

0 = 60.0 J/K + deltaS(B->C)

So the entropy change going from state B to C is has the same magnitude by opposite sign as the entropy change in going from state A to B.

deltaS(B->C) = -60.0 J/K

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