91Ó°ÊÓ

An isothermal process is when the temperature (T) remains constant during expansion or compression. During an adiabatic process, there is no exchange of heat (Q), and thus all the work done on the system affects the internal energy.

For an isothermal process, the final temperature (Tf_iso) remains the same as the initial temperature (Ti), because the temperature is constant during the process. For the adiabatic process, we need to look at the relation between the initial and final temperature: \(PV^{\gamma} = constant\), where \(\gamma\) is the adiabatic index for the gas Since initial pressure (Pi), volume (Vi), and temperature (Ti) are the same for both gases, we can write: \((P_iV_i)^{\gamma} = (P_fV_f)^{\gamma}\) Now we will use the Ideal Gas Law to find the final temperature (Tf_ad) for the adiabatic process: \(P_iV_i = nRT_i\) \(P_fV_f = nRT_f\) Dividing the equations gives us: \(\frac{P_i}{P_f} = \frac{T_f}{T_i}\) Substitute Tf_iso and Tf_ad for Tf in the previous equation, and we get: \(T_f = T_i\) for the isothermal process. \(T_f < T_i\) for the adiabatic process. So the final temperature is greater for the isothermal process (Option A).

For the isothermal process, the work done (W_iso) can be calculated: \(W_{iso} = nRT_i \ln(\frac{V_f}{V_i})\) For the adiabatic process, the work done (W_ad) can be calculated: \(W_{ad} = \frac{nRT_i}{\gamma - 1}(1 - \frac{P_f}{P_i})\) Since both processes have the same initial and final volumes, we can compare the two equations: \(W_{iso} = nRT_i \ln(\frac{V_f}{V_i}) > \frac{nRT_i}{\gamma - 1}(1 - \frac{P_f}{P_i}) = W_{ad}\) So the work done by the gas is greater for the isothermal process (Option C). Therefore, the correct answer is (A) and (C).