{"id":42354,"date":"2025-06-29T02:47:42","date_gmt":"2025-06-29T02:47:42","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=42354"},"modified":"2025-06-29T02:47:45","modified_gmt":"2025-06-29T02:47:45","slug":"the-atmospheric-pressure-in-denver-colorado-is-usually-about-84-0-kpa","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/the-atmospheric-pressure-in-denver-colorado-is-usually-about-84-0-kpa\/","title":{"rendered":"The atmospheric pressure in Denver, Colorado, is usually about 84.0 kPa"},"content":{"rendered":"\n<p>The atmospheric pressure in Denver, Colorado, is usually about 84.0 kPa. What is this pressure in atm and torr units? 71. At an ocean depth of 250 feet, the pressure is about 8.4 atm. Convert the pressure to mm Hg and kPa units.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Let&#8217;s solve the two problems one by one.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">1. Atmospheric pressure in Denver (84.0 kPa) to atm and torr:<\/h3>\n\n\n\n<p>To convert pressure units, we need to use the following relationships:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>1 atm = 101.3 kPa<\/strong><\/li>\n\n\n\n<li><strong>1 atm = 760 torr<\/strong><\/li>\n<\/ul>\n\n\n\n<h4 class=\"wp-block-heading\">Conversion from kPa to atm:<\/h4>\n\n\n\n<p>To convert from kilopascals (kPa) to atmospheres (atm), we use the formula:Pressure&nbsp;in&nbsp;atm=Pressure&nbsp;in&nbsp;kPa101.3\\text{Pressure in atm} = \\frac{\\text{Pressure in kPa}}{101.3}Pressure&nbsp;in&nbsp;atm=101.3Pressure&nbsp;in&nbsp;kPa\u200b<\/p>\n\n\n\n<p>Substituting the given pressure (84.0 kPa):Pressure&nbsp;in&nbsp;atm=84.0101.3\u22480.829\u2009atm\\text{Pressure in atm} = \\frac{84.0}{101.3} \\approx 0.829 \\, \\text{atm}Pressure&nbsp;in&nbsp;atm=101.384.0\u200b\u22480.829atm<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Conversion from kPa to torr:<\/h4>\n\n\n\n<p>Next, to convert from kPa to torr, we use the relationship:Pressure&nbsp;in&nbsp;torr=Pressure&nbsp;in&nbsp;kPa\u00d7760101.3\\text{Pressure in torr} = \\frac{\\text{Pressure in kPa} \\times 760}{101.3}Pressure&nbsp;in&nbsp;torr=101.3Pressure&nbsp;in&nbsp;kPa\u00d7760\u200b<\/p>\n\n\n\n<p>Substituting the given pressure:Pressure&nbsp;in&nbsp;torr=84.0\u00d7760101.3\u2248628.3\u2009torr\\text{Pressure in torr} = \\frac{84.0 \\times 760}{101.3} \\approx 628.3 \\, \\text{torr}Pressure&nbsp;in&nbsp;torr=101.384.0\u00d7760\u200b\u2248628.3torr<\/p>\n\n\n\n<p>So, the pressure in Denver is approximately:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>0.829 atm<\/strong><\/li>\n\n\n\n<li><strong>628.3 torr<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">2. Ocean depth pressure of 8.4 atm to mm Hg and kPa:<\/h3>\n\n\n\n<p>Now, let&#8217;s convert the pressure at an ocean depth of 250 feet (8.4 atm) to mm Hg and kPa.<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Conversion from atm to mm Hg:<\/h4>\n\n\n\n<p>Using the relationship:1\u2009atm=760\u2009mm&nbsp;Hg1 \\, \\text{atm} = 760 \\, \\text{mm Hg}1atm=760mm&nbsp;Hg<\/p>\n\n\n\n<p>We multiply the given pressure in atm by 760:Pressure&nbsp;in&nbsp;mm&nbsp;Hg=8.4\u00d7760=6384\u2009mm&nbsp;Hg\\text{Pressure in mm Hg} = 8.4 \\times 760 = 6384 \\, \\text{mm Hg}Pressure&nbsp;in&nbsp;mm&nbsp;Hg=8.4\u00d7760=6384mm&nbsp;Hg<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Conversion from atm to kPa:<\/h4>\n\n\n\n<p>To convert from atm to kPa, we use:1\u2009atm=101.3\u2009kPa1 \\, \\text{atm} = 101.3 \\, \\text{kPa}1atm=101.3kPa<\/p>\n\n\n\n<p>So:Pressure&nbsp;in&nbsp;kPa=8.4\u00d7101.3=850.9\u2009kPa\\text{Pressure in kPa} = 8.4 \\times 101.3 = 850.9 \\, \\text{kPa}Pressure&nbsp;in&nbsp;kPa=8.4\u00d7101.3=850.9kPa<\/p>\n\n\n\n<p>Thus, the pressure at a depth of 250 feet is approximately:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>6384 mm Hg<\/strong><\/li>\n\n\n\n<li><strong>850.9 kPa<\/strong><\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The atmospheric pressure in Denver (84.0 kPa) is about <strong>0.829 atm<\/strong> and <strong>628.3 torr<\/strong>.<\/li>\n\n\n\n<li>The pressure at an ocean depth of 250 feet (8.4 atm) is about <strong>6384 mm Hg<\/strong> and <strong>850.9 kPa<\/strong>.<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"1024\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-306.jpeg\" alt=\"\" class=\"wp-image-42355\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-306.jpeg 1024w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-306-300x300.jpeg 300w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-306-150x150.jpeg 150w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner5-306-768x768.jpeg 768w\" sizes=\"auto, (max-width: 1024px) 100vw, 1024px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The atmospheric pressure in Denver, Colorado, is usually about 84.0 kPa. What is this pressure in atm and torr units? 71. At an ocean depth of 250 feet, the pressure is about 8.4 atm. Convert the pressure to mm Hg and kPa units. The Correct Answer and Explanation is: Let&#8217;s solve the two problems one [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-42354","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42354","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=42354"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42354\/revisions"}],"predecessor-version":[{"id":42356,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/42354\/revisions\/42356"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=42354"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=42354"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=42354"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}