{"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":"9414c161-8a90-5060-8722-0f83e6eaacf7","task_key":"default--test--4271","task_revision_id":"1","upstream_id":"4271","short_description":"Assume all gases are perfect unless stated otherwise. Note that 1 atm = 1.013 25…","config":"default","split":"test","body":"{\"category\":\"chemistry\",\"options\":\"['29.80$\\\\\\\\mathrm{J} \\\\\\\\mathrm{K}^{-1} \\\\\\\\mathrm{~mol}^{-1}$', '47.80$\\\\\\\\mathrm{J} \\\\\\\\mathrm{K}^{-1} \\\\\\\\mathrm{~mol}^{-1}$', '38.70$\\\\\\\\mathrm{J} \\\\\\\\mathrm{K}^{-1} \\\\\\\\mathrm{~mol}^{-1}$', '44.20$\\\\\\\\mathrm{J} \\\\\\\\mathrm{K}^{-1} \\\\\\\\mathrm{~mol}^{-1}$', '55.00$\\\\\\\\mathrm{J} \\\\\\\\mathrm{K}^{-1} \\\\\\\\mathrm{~mol}^{-1}$', '33.60$\\\\\\\\mathrm{J} \\\\\\\\mathrm{K}^{-1} \\\\\\\\mathrm{~mol}^{-1}$', '41.40$\\\\\\\\mathrm{J} \\\\\\\\mathrm{K}^{-1} \\\\\\\\mathrm{~mol}^{-1}$', '50.60$\\\\\\\\mathrm{J} \\\\\\\\mathrm{K}^{-1} \\\\\\\\mathrm{~mol}^{-1}$', '35.20$\\\\\\\\mathrm{J} \\\\\\\\mathrm{K}^{-1} \\\\\\\\mathrm{~mol}^{-1}$', '62.50$\\\\\\\\mathrm{J} \\\\\\\\mathrm{K}^{-1} \\\\\\\\mathrm{~mol}^{-1}$']\",\"question\":\"Assume all gases are perfect unless stated otherwise. Note that 1 atm = 1.013 25 bar. Unless otherwise stated, thermochemical data are for 298.15 K. The constant-volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabatically and reversibly. If the decrease in pressure is also measured, we can use it to infer the value of $\\\\gamma=C_p / C_V$ and hence, by combining the two values, deduce the constant-pressure heat capacity. A fluorocarbon gas was allowed to expand reversibly and adiabatically to twice its volume; as a result, the temperature fell from $298.15 \\\\mathrm{~K}$ to $248.44 \\\\mathrm{~K}$ and its pressure fell from $202.94 \\\\mathrm{kPa}$ to $81.840 \\\\mathrm{kPa}$. Evaluate $C_p$.\",\"src\":\"scibench-atkins\"}","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":[]}