Final answer:
In an isothermal expansion, the heat absorbed can be found using the ideal gas law. The heat transferred in the described process can be calculated by considering two separate steps. In the quasi-static transformation, the heat transferred is zero due to it being isothermal.
Step-by-step explanation:
In order to solve this problem, we need to use the ideal gas law and 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
(a) To find the heat absorbed during the isothermal expansion, we can use the formula:
Q = nRT ln(Vf/Vi)
where Q is the heat absorbed, n is the number of moles of gas, R is the gas constant, T is the initial temperature, and Vf and Vi are the final and initial volumes respectively.
Plugging in the given values, we have:
Q = (1 mol)(8.31 J/mol·K)(300 K) ln(2) ≈ 1447 J
(b) To find the heat transferred in the described process, we need to consider two separate steps: first, the decrease in pressure at constant volume, which involves no heat transfer as it is adiabatic; and second, the isobaric expansion, for which we can use the formula:
Q = nCpΔT
where Q is the heat transferred, n is the number of moles of gas, Cp is the molar heat capacity at constant pressure, and ΔT is the temperature change.
Given that it is isothermal, the temperature change is zero, so the heat transferred for this case is also zero.
(c) To calculate the heat transfer when the gas is transformed quasi-statically, we use the same formula as in part (b) for the isobaric expansion, with the difference that there is also a decrease in pressure at constant volume before the expansion:
Q = nCpΔT
Again, since it is isothermal, the change in temperature is zero, so the heat transfer in this case is also zero.