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Correct answer: The pressure at B is 2.0 bar
A->B isothermal: T_B = 500 K. B->C isochoric: P_B/T_B = P_C/T_C => P_B = 1.0*(500/250) = 2.0 bar. (A correct.) C->D irreversible adiabatic: -P_ext(V_D - V_C) = nCv(T_D - T_C). Substituting V = nRT/P: -(3)[nRT_D/3 - nRT_C/1] = n(3/2)R(T_D - 250). => -(T_D - 750) = (3/2)(T_D - 250) => -T_D + 750 = (3/2)T_D - 375 => 1125 = (5/2)T_D => T_D = 450 K. (B correct.) Delta_H_CD = nCp(T_D - T_C) = 2*(5/2)R*(200) = 1000R. (C correct.) Delta_U_BC = nCv(T_C - T_B) = 2*(3/2)R*(250-500) = -750R. Option D states +375R which is wrong. (D incorrect.)