{"id":1437,"date":"2018-03-07T17:46:20","date_gmt":"2018-03-07T09:46:20","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1437"},"modified":"2018-03-07T17:46:20","modified_gmt":"2018-03-07T09:46:20","slug":"boyles-law-constant-temperature-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/boyles-law-constant-temperature-problems-and-solutions.htm","title":{"rendered":"Boyle’s law (constant temperature) – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. <\/span><\/span>Some ideal gases initially have pressure<\/a> P and volume V. If the gas undergoes isothermal process<\/a> so that the final pressure becomes 4 times the initial pressure, then the final volume of gas is…<\/span><\/span><\/span><\/p>\n

Known :<\/span><\/span><\/u><\/p>\n

Initial pressure <\/span><\/span>(P<\/span><\/span>1<\/span><\/sub>) = P<\/span><\/span><\/p>\n

Final pressure <\/span><\/span>(P<\/span><\/span>2<\/span><\/sub>) = 4P<\/span><\/span><\/p>\n

Initial volume <\/span><\/span>(V<\/span><\/span>1<\/span><\/sub>) = V<\/span><\/span><\/p>\n

Wanted:<\/u><\/span><\/span> Final volume <\/span><\/span>(V<\/span><\/span>2<\/span><\/sub>)<\/span><\/span><\/p>\n

Solution :<\/span><\/span><\/u><\/p>\n

The formula of Boyle’s law<\/a> :<\/span><\/span><\/p>\n

P V = <\/span><\/span>constant<\/span><\/span><\/p>\n

P<\/span><\/span>1<\/span><\/sub> V<\/span><\/span>1<\/span><\/sub> = P<\/span><\/span>2<\/span><\/sub> V<\/span><\/span>2<\/span><\/sub><\/p>\n

(P)(V) = (4P)(V<\/span><\/span>2<\/span><\/sub>)<\/span><\/span><\/p>\n

V = 4 V<\/span><\/span>2<\/span><\/sub><\/p>\n

V<\/span><\/span>2 <\/span><\/sub>= V \/ 4 = \u00bc V<\/span><\/span><\/p>\n

The final volume of gases is <\/span>\u00bc <\/span>times the initial volume.<\/span><\/span><\/p>\n\n

2. In a closed container, the gas expands so that the <\/span>final <\/span>volume <\/span>becomes <\/span>2 times the initial volume (V = initial volume, P = initial pressure). <\/span>The final pressure of gases is…<\/span><\/span><\/span><\/p>\n

Known :<\/span><\/u><\/span><\/p>\n

Initial pressure <\/span>(P<\/span>1<\/span><\/sub>) = P<\/span><\/span><\/p>\n

Initial volume <\/span>(V<\/span>1<\/span><\/sub>) = V<\/span><\/span><\/p>\n

Final volume <\/span>(V<\/span>2<\/span><\/sub>) = 2V<\/span><\/span><\/p>\n

Wanted <\/u><\/span>:<\/u><\/span> Final pressure <\/span>(P<\/span>2<\/span><\/sub>)<\/span><\/span><\/p>\n

Solution :<\/span><\/u><\/span><\/p>\n

P<\/span>1<\/span><\/sub> V<\/span>1<\/span><\/sub> = P<\/span>2<\/span><\/sub> V<\/span>2<\/span><\/sub><\/span><\/p>\n

P V = <\/span>P<\/span>2<\/span><\/sub> (2V)<\/span><\/span><\/p>\n

P = <\/span>P<\/span>2<\/span><\/sub> (2)<\/span><\/span><\/p>\n

P<\/span>2 <\/span><\/sub>= P \/ 2 = \u00bd P<\/span><\/span><\/p>\n

The gases pressure becomes \u00bd times the initial pressure. <\/span><\/span><\/p>\n\n

3. In a closed container, gases having a pressure of 2 atm and a volume of 1 liter. If gas pressure becomes 4 atm then gas volume becomes \u2026<\/span><\/span><\/span><\/span><\/p>\n

Known :<\/span><\/u><\/span><\/span><\/span><\/p>\n

Initial pressure <\/span>(P<\/span>1<\/span><\/sub>) = 2 atm = 2 x 10<\/span>5<\/span><\/sup> Pa<\/span><\/span><\/span><\/span><\/p>\n

Final pressure <\/span>(P<\/span>2<\/span><\/sub>) = 4 atm = 4 x 10<\/span>5<\/span><\/sup> Pa<\/span><\/span><\/span><\/span><\/p>\n

Initial volume <\/span>(V<\/span>1<\/span><\/sub>) = 1 liter = 1 dm<\/span>3<\/span><\/sup> = 1 x 10<\/span>-3<\/span><\/sup> m<\/span>3<\/span><\/sup><\/span><\/span><\/span><\/p>\n

Wanted<\/u><\/span> : <\/span>Final volume <\/span>(V<\/span>2<\/span><\/sub>)<\/span><\/span><\/span><\/span><\/p>\n

Solution :<\/span><\/u><\/span><\/span><\/span><\/p>\n

P<\/span><\/span>1<\/span><\/sub> V<\/span><\/span>1<\/span><\/sub> = P<\/span><\/span>2<\/span><\/sub> V<\/span><\/span>2<\/span><\/sub><\/p>\n

(<\/span><\/span>2 x 10<\/span>5<\/span><\/sup>)(1 x 10<\/span>-3<\/span><\/sup>) = (4 x 10<\/span>5<\/span><\/sup>)<\/span> V<\/span><\/span>2<\/span><\/sub><\/p>\n

(<\/span><\/span>1)(1 x 10<\/span>-3<\/span><\/sup>) = (2)<\/span> V<\/span><\/span>2<\/span><\/sub><\/p>\n

1 x 10<\/span>-3<\/span><\/sup> = (2)<\/span> V<\/span><\/span>2<\/span><\/sub><\/p>\n

V<\/span>2 <\/span><\/sub>= \u00bd x <\/span>10<\/span>-3<\/span><\/sup> <\/span><\/span><\/span><\/p>\n

V<\/span>2<\/span><\/sub> = 0.5 x 10<\/span>-3 <\/span><\/sup>m<\/span>3<\/span><\/sup> = 0.5 dm<\/span>3<\/span><\/sup> = 0.5 liter<\/span>s<\/span> <\/span><\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

1. Some ideal gases initially have pressure P and volume V. If the gas undergoes isothermal process so that the final pressure becomes 4 times the initial pressure, then the final volume of gas is… Known : Initial pressure (P1) = P Final pressure (P2) = 4P Initial volume (V1) = V Wanted: Final volume … Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_seopress_titles_title":"","_seopress_titles_desc":"","_seopress_robots_index":"","_seopress_robots_follow":"","_seopress_robots_imageindex":"","_seopress_robots_snippet":"","_seopress_robots_primary_cat":"","_seopress_robots_breadcrumbs":"","_seopress_robots_freeze_modified_date":"","_seopress_robots_custom_modified_date":"","_seopress_robots_canonical":"","_seopress_social_fb_title":"","_seopress_social_fb_desc":"","_seopress_social_fb_img":"","_seopress_social_fb_img_attachment_id":0,"_seopress_social_fb_img_width":0,"_seopress_social_fb_img_height":0,"_seopress_social_twitter_title":"","_seopress_social_twitter_desc":"","_seopress_social_twitter_img":"","_seopress_social_twitter_img_attachment_id":0,"_seopress_social_twitter_img_width":0,"_seopress_social_twitter_img_height":0,"_seopress_redirections_value":"","_seopress_redirections_enabled":"","_seopress_redirections_enabled_regex":"","_seopress_redirections_logged_status":"","_seopress_redirections_param":"","_seopress_redirections_type":0,"_seopress_analysis_target_kw":"Boyle's law (constant temperature) - problems and solutions","_seopress_news_disabled":"","_seopress_video_disabled":"","_seopress_video":[],"_seopress_pro_schemas_manual":[],"_seopress_pro_rich_snippets_disable_all":"","_seopress_pro_rich_snippets_disable":[],"_seopress_pro_schemas":[],"footnotes":""},"categories":[3],"tags":[],"class_list":["post-1437","post","type-post","status-publish","format-standard","hentry","category-solved-problems-in-basic-physics"],"gt_translate_keys":[{"key":"link","format":"url"}],"_links":{"self":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1437","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/comments?post=1437"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1437\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1437"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1437"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1437"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}