# Suppose 4 mol of an ideal gas undergo a reversible isothermal expansion from volume V1 to volume V2 = 2V1 at temperature T=400 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 charge of the gas.

Arun
25757 Points
4 years ago
The differential work done by expansion against a pressure, P is given by:

δw = P dV

For an ideal gas

P = n*R*T/V, so:

δw_rev = n*R*T/V dV

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

w_rev = n*R*T*ln(V_final/V_initial)

Here, n = 4 moles, T = 400K, and V_final = 2*V_initial, so:

w_rev = (4mol)*(8.314 J/(mol*K))*(400K)*ln(2)

w_rev = 9221 J.

This is the work done by the gas on the surroundings.

There are several ways to think about what the entropy change must be. Perhaps the easiest is to recognize that the internal energy of an ideal gas depends only on temperature. In this case, the process is isothermal, so the internal energy of the gas doesn't change. From the first law, we can then write:

ΔE = 0 = q - w

where q is the thermal energy added to the gas.

q = w.

But ΔS_rev = q_rev/T, so:

ΔS_rev = (9221 J)/(400K) = 23.05 J/K

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.)