The internal energy of a monatomic ideal gas is given by the equation:
U = (3/2) nRT
where U is the internal energy, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.
At the initial state, the gas is at a temperature of T0 and a pressure of P0. The number of moles is given as one, so we have:
U0 = (3/2) RT0
To find the final internal energy and the work done by the gas, we need to know whether the heat is added at constant pressure or constant volume. Let's consider each case separately:
Case 1: Heat is added at constant pressure
If the heat is added at constant pressure, then the change in internal energy is given by:
ΔU = q - PΔV
where q is the heat added, P is the constant pressure, and ΔV is the change in volume.
Since the gas is monatomic, its molar specific heat at constant pressure is Cp = (5/2)R. Thus, the heat added at constant pressure is given by:
q = nCpΔT
where ΔT is the change in temperature.
Since the pressure is constant, we have:
ΔV = nRΔT/P
Substituting these expressions into the equation for ΔU, we get:
ΔU = nCpΔT - PΔV
ΔU = nCpΔT - nRΔT
ΔU = n(Cp - R)ΔT
The work done by the gas is given by:
W = -PΔV
Substituting ΔV from above, we get:
W = -nRΔT
The final internal energy is given by:
Uf = U0 + ΔU
Uf = U0 + n(Cp - R)ΔT
Uf = (3/2) RT0 + (5/2)RΔT
Case 2: Heat is added at constant volume
If the heat is added at constant volume, then the change in internal energy is given by:
ΔU = q
Since the volume is constant, the gas does no work, so W = 0.
The heat added at constant volume is given by:
q = nCvΔT
where Cv is the molar specific heat at constant volume, which for a monatomic gas is Cv = (3/2)R.
Substituting these expressions into the equation for ΔU, we get:
ΔU = nCvΔT
ΔU = (3/2) RT0
The final internal energy is given by:
Uf = U0 + ΔU
Uf = (3/2) RT0 + (3/2) RT0
Uf = 3 RT0
Therefore, the initial internal energy of the gas is (3/2) RT0. The final internal energy and work done by the gas depend on whether the heat is added at constant pressure or constant volume. If the heat is added at constant pressure, then the final internal energy is given by (3/2) RT0 + (5/2)RΔT, and the work done by the gas is -nRΔT. If the heat is added at constant volume, then the final internal energy is given by 3 RT0, and the work done by the gas is 0.