Final answer:
In an isobaric process, the change in internal energy is equal to the heat transferred and the work done. For an isochoric process, the work done is zero. The heat transferred can be calculated using the ideal gas law and the molar heat capacity.
Step-by-step explanation:
In an isobaric process, the pressure remains constant, so the change in internal energy (ΔU) is equal to the heat transferred (q) and the work done (w). To calculate the change in internal energy, use the equation ΔU = q + w. In this case, since the pressure is constant, the work done can be calculated using the equation w = PΔV. The heat transferred can be calculated using the equation q = nCpΔT, where n is the number of moles, Cp is the molar heat capacity at constant pressure, and ΔT is the change in temperature.
For part A) of the question:
Given: P = 2 atm, V1 = 2L, V2 = 4L, n = 1 mole
Since it is an isobaric process, ΔU = q + w. The work done can be calculated using w = PΔV = (2 atm)(4L - 2L) = 4 atm*L.
The heat transferred can be calculated using q = nCpΔT = (1 mole)(Cp)(ΔT). Since the temperature is constant in an isobaric process, ΔT = 0, so q = 0.
Therefore, ΔU = 0 + 4 atm*L = 4 atm*L.
For part B) of the question:
Given: V = 2L, P1 = 2 atm, P2 = 4 atm, n = 1 mole
Since it is an isochoric process, the volume remains constant, so the work done (w) is zero.
The heat transferred can be calculated using q = nCpΔT = (1 mole)(Cp)(ΔT). Since the pressure is constant in an isochoric process, ΔT can be calculated using the ideal gas law: P1V1/T1 = P2V2/T2. Rearranging the equation to solve for ΔT gives ΔT = (P2V2 - P1V1)/(nR). Substituting the given values into the equation, ΔT = (4 atm)(2 L - 2 L)/(1 mole * R), where R is the ideal gas constant. Note that the units must be consistent.
Therefore, ΔU = q + w = (1 mole)(Cp)(ΔT) + 0 = (1 mole)(Cp)((4 atm)(2 L - 2 L)/(1 mole * R)).