{"id":1470,"date":"2018-03-08T13:20:11","date_gmt":"2018-03-08T05:20:11","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1470"},"modified":"2018-03-08T13:20:11","modified_gmt":"2018-03-08T05:20:11","slug":"ideal-gas-law-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/ideal-gas-law-problems-and-solutions.htm","title":{"rendered":"Ideal gas law – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

1. I<\/span><\/span>deal gases in a <\/span><\/span><\/span>closed container <\/span><\/span><\/span>initially have volume V and temperature<\/a> T. The final temperature is 5\/4T and the final pressure<\/a> is 2P. What is the final volume of the gas? <\/span><\/span><\/span><\/p>\n

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

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

Initial temperature (T<\/span><\/span>1<\/span><\/sub>) = T<\/span><\/span><\/p>\n

Final temperature (T<\/span><\/span>2<\/span><\/sub>) = 5\/4 T<\/span><\/span><\/p>\n

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

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

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

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

\"Ideal<\/p>\n

<\/p>\n

2. Determine the volume of 2.00 mol<\/span><\/span>es<\/span><\/span> of gases (ideal gas) at STP. STP = Standard Temperature and Pressure. <\/span><\/span><\/p>\n

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

Moles of gas (n) = 2 mol<\/span><\/span>es<\/span><\/span><\/p>\n

Standard temperature (T) = 0 <\/span><\/span>o<\/span><\/span><\/sup>C = 0 + 273 = 273 Kelvin<\/span><\/span><\/p>\n

Standard pressure (P) = 1 atm = 1.013 x 10<\/span><\/span>5<\/span><\/span><\/sup> Pa<\/span><\/span><\/p>\n

Universal gas constant (R) = 8.315 Joule\/mol<\/span><\/span>e<\/span><\/span>.Kelvin<\/span><\/span><\/p>\n

Wanted <\/u><\/span><\/span>: Volume of gases (V)<\/span><\/span><\/p>\n

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

Equation of Ideal gas law <\/a><\/b><\/u><\/span><\/span>(in the number of moles, n)<\/u><\/span><\/span><\/p>\n

\"Ideal<\/p>\n

Volume 2 moles of gases is 44.8 liters.<\/span><\/span><\/p>\n

Volume 1 mol of gases is 45.4 liters \/ 2 = 22.4 liters.<\/span><\/span><\/p>\n

Volume 1 mol of any gases is 22.4 liters. <\/span><\/span><\/p>\n

<\/p>\n

3. 4 liters of oxygen gas has a temperature of 27\u00b0C and pressure of 2 atm (1 atm = 10<\/span>5<\/span><\/sup> Pa) in a closed container. Universal gas constant (R) = 8.314 J.mol<\/span>e<\/span>\u22121<\/span><\/sup>.K<\/span>\u22121<\/span><\/sup> and Avogadro’s number (<\/span>N<\/span>A<\/span><\/sub>) = 6.02<\/span> x 10<\/span>23<\/span><\/sup> molecules\/mol<\/span>e<\/span>. What are the molecules of oxygen gases in the container? <\/span><\/span><\/span><\/span><\/p>\n

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

Volume of gases (V) = 4 liters = 4 dm<\/span><\/span>3<\/span><\/span><\/sup> = 4 x 10<\/span><\/span>-3<\/span><\/span><\/sup> m<\/span><\/span>3<\/span><\/span><\/sup><\/p>\n

Temperature of gases (T) = 27<\/span><\/span>o<\/span><\/span><\/sup>C = 27 + 273 = 300 Kelvin <\/span><\/span><\/p>\n

Pressure of gases (P) = 2 atm = 2 x 10<\/span><\/span>5 <\/span><\/span><\/sup>Pa<\/span><\/span><\/p>\n

Universal gas constant (R) = 8.314 J.mol<\/span><\/span>e<\/span><\/span>\u22121<\/span><\/span><\/sup>.K<\/span><\/span>\u22121<\/span><\/span><\/sup><\/p>\n

Avogadro’s number (N<\/span><\/span>A<\/span><\/sub>) = 6.02 x 10<\/span><\/span>23<\/span><\/span><\/sup><\/p>\n

Wanted <\/u><\/span><\/span>: What is the molecules of oxygen gases in the container (N)<\/span><\/span><\/p>\n

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

\"Ideal<\/p>\n

In 1 mole oxygen gases, there are 1.93 x 10<\/span><\/span>23<\/span><\/span><\/sup> oxygen molecules.<\/span><\/span><\/p>\n

<\/p>\n

4. <\/span><\/span>A container containing a neon gas (Ne, atomic mass = 20 u) at standard temperature and pressure (STP<\/span><\/span><\/span>) <\/span><\/span><\/span>has a volume of 2 m<\/span><\/span><\/span>3<\/span><\/span><\/span><\/sup>. Determine the mass of the neon gas!<\/span><\/span><\/span><\/p>\n

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

Atomic mass of neon <\/span><\/span>= 20 gram\/mol<\/span><\/span>e<\/span><\/span> = 0,02 kg\/mol<\/span><\/span>e<\/span><\/span><\/p>\n

Standard temperature <\/span><\/span>(T) = 0<\/span><\/span>o<\/span><\/span><\/sup>C = 273 Kelvin<\/span><\/span><\/p>\n

Standard pressure <\/span><\/span>(P) = 1 atm = 1.013 x 10<\/span><\/span>5<\/span><\/span><\/sup> Pascal<\/span><\/span><\/p>\n

Volume (V) = 2 m<\/span><\/span>3<\/span><\/span><\/sup><\/p>\n

Wanted <\/u><\/span><\/span>:<\/u><\/span><\/span> mass <\/span><\/span>(m) <\/span><\/span>of neon gas<\/span><\/span><\/p>\n

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

At standard temperature and pressure <\/span><\/span>(STP), 1 mol<\/span><\/span>e<\/span><\/span> of any gases<\/span><\/span>, <\/span><\/span>include neon gas<\/span><\/span>, <\/span><\/span>have volume <\/span><\/span>22.4 liter<\/span><\/span>s<\/span><\/span> = 22.4 dm<\/span><\/span>3<\/span><\/span><\/sup> = 0.0448 m<\/span><\/span>3<\/span><\/span><\/sup>. <\/span><\/span><\/p>\n

\"Ideal<\/p>\n

In the volume of 2 m<\/span><\/span>3<\/span><\/span><\/sup>,\u00a0<\/span><\/span>there are 44.6 moles of neon gas.<\/span><\/span><\/p>\n

Relative atomic mass of neon gas is 20 gram\/mole. <\/span><\/span><\/p>\n

This means that in 1 mole there are 20 grams or 0.02 kg of neon gas. Because in 1 mol there are 0.02 kg of neon gas then in 44.6 mole there are 44.6 moles x 0.02 kg\/mole = 0.892 kg = 892 gram of neon gases.<\/span><\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

1. Ideal gases in a closed container initially have volume V and temperature T. The final temperature is 5\/4T and the final pressure is 2P. What is the final volume of the gas? Known : Initial volume (V1) = V Initial temperature (T1) = T Final temperature (T2) = 5\/4 T Initial pressure (P1) = … 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":"Ideal gas law - 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-1470","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\/1470","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=1470"}],"version-history":[{"count":0,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1470\/revisions"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1470"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1470"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1470"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}