We can solve this problem by using the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:
ΔU = Q - W
Since the container is well-insulated, Q = 0 and therefore ΔU = -W. The change in internal energy is given by:
ΔU = (3/2)nRΔT
where n is the number of moles of gas, R is the gas constant, and ΔT is the change in temperature. Since the gas is monatomic, we can substitute n = N/NA, where N is the number of atoms and NA is Avogadro's number, and use R = kNA, where k is the Boltzmann constant. Then we have:
ΔU = (3/2)(N/NA)kΔT
The work done by the gas is given by:
W = PextΔV
where Pext is the external pressure and ΔV is the change in volume. Since the pressure is constant, we can substitute Pext = Patm, the atmospheric pressure. Then we have:
W = Patm(V2 - V1)
where V1 and V2 are the initial and final volumes, respectively. Substituting the given values, we have:
W = 5 J
V1 = 16 cc = 16×10^-6 m^3
V2 = 3 cc = 3×10^-6 m^3
ΔV = V2 - V1 = -13×10^-6 m^3 (negative because the gas is compressed)
Substituting into the work equation, we get:
5 J = (101325 Pa)(-13×10^-6 m^3)
P = -5/(101325×13×10^-6) atm
P ≈ 0.003 atm
This result is negative, which means that the gas has done work on the surroundings rather than the other way around. This is because we have compressed the gas by doing work on it, and the gas has then expanded against the walls of the container, doing work on the surroundings. To get the final pressure of the gas, we need to add the atmospheric pressure to the pressure change caused by the compression:
Pf = Patm - ΔP = Patm - W/V2 = 1 - 5/(3×10^6) atm
Pf ≈ 0.9983 atm
Therefore, the final pressure of the gas is 0.9983 atm.