{"id":233727,"date":"2025-06-13T11:01:11","date_gmt":"2025-06-13T11:01:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=233727"},"modified":"2025-06-13T11:01:14","modified_gmt":"2025-06-13T11:01:14","slug":"a-weather-balloon-rises-through-the-atmosphere","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/13\/a-weather-balloon-rises-through-the-atmosphere\/","title":{"rendered":"A weather balloon rises through the atmosphere"},"content":{"rendered":"\n<p>A weather balloon rises through the atmosphere, its volume expanding from<br>to20 to 10<br>as the temperature drops from<br>to 1.0<br>If the initial gas pressure inside the balloon is 4.0 to12m<br>what is the final pressure<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we apply the <strong>Combined Gas Law<\/strong>, which relates the pressure, volume, and temperature of a gas:P1V1T1=P2V2T2\\frac{P_1 V_1}{T_1} = \\frac{P_2 V_2}{T_2}T1\u200bP1\u200bV1\u200b\u200b=T2\u200bP2\u200bV2\u200b\u200b<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>P1P_1P1\u200b, V1V_1V1\u200b, and T1T_1T1\u200b are the <strong>initial pressure<\/strong>, <strong>volume<\/strong>, and <strong>temperature<\/strong>,<\/li>\n\n\n\n<li>P2P_2P2\u200b, V2V_2V2\u200b, and T2T_2T2\u200b are the <strong>final pressure<\/strong>, <strong>volume<\/strong>, and <strong>temperature<\/strong>.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Write down known values<\/h3>\n\n\n\n<p>Let\u2019s assume your incomplete numbers were meant to be:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial volume: V1=2.0\u2009LV_1 = 2.0 \\, \\text{L}V1\u200b=2.0L<\/li>\n\n\n\n<li>Final volume: V2=10.0\u2009LV_2 = 10.0 \\, \\text{L}V2\u200b=10.0L<\/li>\n\n\n\n<li>Initial temperature: T1=20\u2218C=293\u2009KT_1 = 20^\\circ\\text{C} = 293 \\, \\text{K}T1\u200b=20\u2218C=293K<\/li>\n\n\n\n<li>Final temperature: T2=1.0\u2218C=274\u2009KT_2 = 1.0^\\circ\\text{C} = 274 \\, \\text{K}T2\u200b=1.0\u2218C=274K<\/li>\n\n\n\n<li>Initial pressure: P1=4.0\u00d7102\u2009mmHgP_1 = 4.0 \\times 10^{2} \\, \\text{mmHg}P1\u200b=4.0\u00d7102mmHg<\/li>\n\n\n\n<li>Final pressure: P2=?P_2 = ?P2\u200b=?<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Rearrange the Combined Gas Law to solve for P2P_2P2\u200b:<\/h3>\n\n\n\n<p>P2=P1V1T2V2T1P_2 = \\frac{P_1 V_1 T_2}{V_2 T_1}P2\u200b=V2\u200bT1\u200bP1\u200bV1\u200bT2\u200b\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Plug in the values:<\/h3>\n\n\n\n<p>P2=(4.0\u00d7102)(2.0)(274)(10.0)(293)P_2 = \\frac{(4.0 \\times 10^2)(2.0)(274)}{(10.0)(293)}P2\u200b=(10.0)(293)(4.0\u00d7102)(2.0)(274)\u200bP2=(800)(274)2930=2192002930\u224874.8\u2009mmHgP_2 = \\frac{(800)(274)}{2930} = \\frac{219200}{2930} \\approx 74.8 \\, \\text{mmHg}P2\u200b=2930(800)(274)\u200b=2930219200\u200b\u224874.8mmHg<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\u2705 Final Answer:<\/h3>\n\n\n\n<p><strong>The final pressure is approximately 74.8\u2009mmHg\\boxed{74.8 \\, \\text{mmHg}}74.8mmHg\u200b.<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udcd8 Explanation (Like in textbooks):<\/h3>\n\n\n\n<p>When a weather balloon rises through the atmosphere, it encounters <strong>lower external pressure and lower temperatures<\/strong>. According to the Combined Gas Law, the relationship between pressure (P), volume (V), and temperature (T) for a fixed amount of gas is:P1V1T1=P2V2T2\\frac{P_1 V_1}{T_1} = \\frac{P_2 V_2}{T_2}T1\u200bP1\u200bV1\u200b\u200b=T2\u200bP2\u200bV2\u200b\u200b<\/p>\n\n\n\n<p>This equation shows that if the temperature decreases and the volume increases, the pressure must decrease. Physically, this makes sense: as the balloon rises and expands due to lower atmospheric pressure, the internal gas cools and exerts less pressure on the balloon walls.<\/p>\n\n\n\n<p>This principle is fundamental in understanding gas behavior in varying atmospheric conditions and is critical in meteorology and high-altitude physics.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-314.jpeg\" alt=\"\" class=\"wp-image-233728\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>A weather balloon rises through the atmosphere, its volume expanding fromto20 to 10as the temperature drops fromto 1.0If the initial gas pressure inside the balloon is 4.0 to12mwhat is the final pressure The Correct Answer and Explanation is: To solve this problem, we apply the Combined Gas Law, which relates the pressure, volume, and temperature [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-233727","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233727","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=233727"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/233727\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=233727"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=233727"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=233727"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}