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The ๐‘ ๐‘‰ -diagram below shows two different processes, A and B, taking an ideal gas in a sealed container from state ๐‘– to state ๐‘“ , whereย  ๐‘ ๐‘“ = ๐‘ ๐‘– . The temperature of the gas when it is in state ๐‘– is [ Select ] A C B the temperature of the gas when it is in state ๐‘“ . greater than equal to less than ย  The work done by the environment on the gas during process A is [ Select ] A B C the work done during process B. greater than equal to less than ย  Compare the magnitude of the heat that flows during both processes - i.e., | ๐‘„ ๐ด | [ Select ] C A B | ๐‘„ ๐ต | . greater than equal to less than ย  Compare the magnitude of the change in thermal energy of the gas during both processes - i.e., | ฮ” ๐ธ th , ๐ด | [ Select ] B C A | ฮ” ๐ธ th , ๐ต | . greater than equal to less than ย 

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To analyze the PV diagram problem, I will step through each comparison in turn and justify every choice with physical reasoning. First comparison (Ti vs Tf): The problem states pf = pi, so the pressure at the initial and final states is the same. For an ideal gas, the temperature is proportional to pV/(nR). Since pi = pf, the comparison of temperatures reduces to comparing the volumes Vi and Vf. On the diagram, the initial state i lies at a larger V than the final state f, meaning Vf < Vi. With the same pressure, a smaller volume corresponds to a smaller temperature, so Tf < Ti. With this in mind: - The statement Ti is greater than Tf is true. - The statement Ti equa......Login to view full explanation

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