Final answer:
In the given two-step process, the temperature decreases during the first step (constant volume cooling), and no work is done. In the second step (constant pressure heating and expansion), the gas absorbs heat and the final temperature is higher than the initial temperature, making statements c) and d) correct.
Step-by-step explanation:
The problem presented involves a two-step process involving an ideal gas, where the gas undergoes changes in pressure, volume, and temperature. In the first step of the problem, the gas is cooled at constant volume until its pressure falls to 1.20 atm. Since the volume does not change, the temperature must decrease for the pressure to decrease, according to Gay-Lussac's law. Hence, statement a) is incorrect as the temperature cannot remain constant if the pressure decreases at constant volume. Statement b) is also incorrect; during a constant volume process, no work is done as the volume does not change, so no work is done on or by the gas in this step.
In the second step, the gas is heated and allowed to expand against a constant external pressure until it reaches 28 L. Since the gas is expanding and doing work against the external pressure, and the expansion is driven by heating, it follows that the gas absorbs heat during this step, making statement d) correct. The fact that the volume increases during this isobaric process (constant pressure) and the gas absorbs heat suggests that the final temperature must be higher after the expansion than the initial 300K, which means that statement c) is also correct.