Question

Suppose 4.00 mol of an ideal gas undergoes a reversible isothermal expansion from volume V1 to volume V2 = 8V1 at temperature T = 300 K. Find (a) the work done by the gas and (b) the entropy change of the gas. (c) If the expansion is reversible and adiabatic instead of isothermal, what is the entropy change of the gas?

Answer #1

a)

The differential work done by expansion against a pressure, P is
given by:

dw = PdV

For an ideal gas

P = nRT/V

dw_{rev} = nRT/ VdV

If the process is carried out isothermally, we can integrate this to obtain:

w_{rev} =
nRT*ln(V_{final}/V_{initial})

Here, n = 4 moles, T = 300K, and V_{final} =
8*V_{initial}, so:

w_{rev} = (4mol) x (8.314 J/(mol*K)) x (300K) x ln(8)

w_{rev} = 20746

b)

q = w.

But deltaS_{rev} = q_{rev}/T, so:

deltaS_{rev} = (20746 J)/(300K) = 69.15 J/K

c)

For part c, not that by definition, for an adiabatic process, q = 0. This is also a reversible process, so the entropy

change of the gas must also be equal to zero. All reversible, adiabatic processes are also isentropic processes.

(If the process is adiabatic but not reversible, then it is not necessarily isentropic.)

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