{"id":19097,"date":"2025-06-13T13:09:33","date_gmt":"2025-06-13T13:09:33","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=19097"},"modified":"2025-06-13T13:09:34","modified_gmt":"2025-06-13T13:09:34","slug":"a-gas-is-at-35-0deg-c-and-2-50-l","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-gas-is-at-35-0deg-c-and-2-50-l\/","title":{"rendered":"A gas is at 35.0\\deg C and 2.50 L"},"content":{"rendered":"\n

A gas is at 35.0\\deg C and 2.50 L. What is the temperature of the gas if the volume is increased to 5.00 L but pressure does not change?<\/p>\n\n\n\n

The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n

To solve this, we use Charles\u2019s Law<\/strong>, which states that for a fixed amount of gas at constant pressure, the volume of the gas is directly proportional to its temperature in Kelvin<\/strong>: V1T1=V2T2\\frac{V_1}{T_1} = \\frac{V_2}{T_2}<\/p>\n\n\n\n

\n\n\n\n

Given:<\/strong><\/h3>\n\n\n\n
    \n
  • Initial temperature, T1=35.0\u2218C=308.15\u2009KT_1 = 35.0^\\circ C = 308.15 \\, K<\/li>\n\n\n\n
  • Initial volume, V1=2.50\u2009LV_1 = 2.50 \\, L<\/li>\n\n\n\n
  • Final volume, V2=5.00\u2009LV_2 = 5.00 \\, L<\/li>\n\n\n\n
  • Final temperature, T2=?T_2 = ?<\/li>\n\n\n\n
  • Pressure remains constant<\/strong><\/li>\n<\/ul>\n\n\n\n
    \n\n\n\n

    Step-by-Step Solution:<\/strong><\/h3>\n\n\n\n

    Rearrange Charles’s Law to solve for T2T_2: T2=T1\u00d7V2V1T_2 = T_1 \\times \\frac{V_2}{V_1} T2=308.15\u2009K\u00d75.00\u2009L2.50\u2009LT_2 = 308.15 \\, K \\times \\frac{5.00 \\, L}{2.50 \\, L} T2=308.15\u00d72=616.30\u2009KT_2 = 308.15 \\times 2 = 616.30 \\, K<\/p>\n\n\n\n

    Convert back to Celsius: T2=616.30\u2212273.15=343.15\u2218CT_2 = 616.30 – 273.15 = \\boxed{343.15^\\circ C}<\/p>\n\n\n\n

    \n\n\n\n

    Explanation (300 Words):<\/strong><\/h3>\n\n\n\n

    When dealing with gases, it\u2019s important to understand how temperature, pressure, and volume relate. One of the foundational gas laws that helps us analyze these relationships is Charles\u2019s Law<\/strong>. This law explains how the volume of a gas increases with increasing temperature<\/strong>, provided the pressure and the amount of gas stay constant.<\/p>\n\n\n\n

    In this problem, a gas is initially at a temperature of 35.0\u00b0C<\/strong> and occupies a volume of 2.50 L<\/strong>. The volume is then increased to 5.00 L<\/strong>, and we\u2019re asked to determine the new temperature assuming the pressure remains constant.<\/p>\n\n\n\n

    First, we must convert temperatures to the Kelvin scale<\/strong>, because gas laws require absolute temperature<\/strong>. This means converting 35.0\u00b0C to 308.15 K<\/strong>. Charles\u2019s Law tells us that volume and temperature are directly proportional: if the volume doubles, the temperature must also double (in Kelvin).<\/p>\n\n\n\n

    Using the formula V1T1=V2T2\\frac{V_1}{T_1} = \\frac{V_2}{T_2}, we find that the final temperature is 616.30 K<\/strong>, which converts to 343.15\u00b0C<\/strong>. This is a significant increase in temperature, which makes sense because doubling the volume at constant pressure requires more energy \u2014 in the form of increased temperature \u2014 to maintain that pressure.<\/p>\n\n\n\n

    This result illustrates how gases expand when heated and how sensitive gas behavior is to temperature changes. It also highlights why heating a sealed gas container can be dangerous: the volume can\u2019t increase, so pressure rises rapidly instead. Charles\u2019s Law gives us a simple but powerful way to predict and understand such changes in real-world and laboratory situations.<\/p>\n","protected":false},"excerpt":{"rendered":"

    A gas is at 35.0\\deg C and 2.50 L. What is the temperature of the gas if the volume is increased to 5.00 L but pressure does not change? The correct answer and explanation is: To solve this, we use Charles\u2019s Law, which states that for a fixed amount of gas at constant pressure, the […]<\/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-19097","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19097","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=19097"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19097\/revisions"}],"predecessor-version":[{"id":19098,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/19097\/revisions\/19098"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=19097"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=19097"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=19097"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}