{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"mmlu-pro","formal_name":"MMLU-Pro","introduction":"MMLU-Pro rebuilds MMLU with ten answer options instead of four and removes items that no longer separate models. Questions span fourteen academic and professional subjects.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://huggingface.co/datasets/TIGER-Lab/MMLU-Pro","indexing_mode":"index","profile":{"resources":[],"task_format":"","scoring":"","metric":"","size":"","answer_access":"","license":"","citation":"","maintainer":"","released":"","why_hard":"","related":[],"history":"","construction":"","limitations":"","saturation":"","scores":[]}},"task_id":"cd27fdbb-8e9a-5255-9aa9-12f719ffbd4c","task_key":"default--test--4409","task_revision_id":"1","upstream_id":"4409","short_description":"The vapor pressure of ethylene is given as a function of temperature by the…","config":"default","split":"test","body":"{\"category\":\"chemistry\",\"options\":\"['12.838 kJ mol^-1', '16.838 kJ mol^-1', '18.838 kJ mol^-1', '19.838 kJ mol^-1', '13.838 kJ mol^-1', '11.838 kJ mol^-1', '15.838 kJ mol^-1', '14.838 kJ mol^-1', '10.838 kJ mol^-1', '17.838 kJ mol^-1']\",\"question\":\"The vapor pressure of ethylene is given as a function of temperature by the equation log P = - [(834.13) / T] + 1.75 log T - 8.375 × 10^-3T + 5.32340 Calculate the enthalpy of vaporization of ethylene at its normal boiling point (- 103.9°C).\",\"src\":\"stemez-PhysicalChemistry\"}","display_format":"text","language":"","answer_status":"published","assets":[],"source_url":"https://huggingface.co/datasets/TIGER-Lab/MMLU-Pro","history":"initial import","indexing_mode":"index","subproblems":[],"grids":[]}