{"id":25853,"date":"2025-06-19T07:07:25","date_gmt":"2025-06-19T07:07:25","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=25853"},"modified":"2025-06-19T07:07:31","modified_gmt":"2025-06-19T07:07:31","slug":"a-sample-of-argon-gas-has-a-volume-of-71-3-ml-at-a-pressure-of-0-763-atm-and-a-temperature-of-37-3-c","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/a-sample-of-argon-gas-has-a-volume-of-71-3-ml-at-a-pressure-of-0-763-atm-and-a-temperature-of-37-3-c\/","title":{"rendered":"A sample of argon gas has a volume of 71.3 mL at a pressure of 0.763 atm and a temperature of 37.3 C"},"content":{"rendered":"\n<p>A sample of argon gas has a volume of 71.3 mL at a pressure of 0.763 atm and a temperature of 37.3 C. What is the final volume of the gas, in mL, when the pressure and temperature are changed to 827 mug and 402 K<\/p>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\"><strong>The correct answer and explanation is:<\/strong><\/mark><\/p>\n\n\n\n<p>To solve this problem, we need to use 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}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Identify known values<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Initial Pressure P1=0.763P_1 = 0.763 atm<\/li>\n\n\n\n<li>Initial Volume V1=71.3V_1 = 71.3 mL<\/li>\n\n\n\n<li>Initial Temperature T1=37.3\u2218C=37.3+273.15=310.45T_1 = 37.3^\\circ C = 37.3 + 273.15 = 310.45 K<\/li>\n\n\n\n<li>Final Pressure P2=827P_2 = 827 \u03bcg (This needs to be converted to <strong>atm<\/strong> or another gas pressure unit. Here, <strong>\u03bcg<\/strong> (micrograms) is a unit of mass, not pressure**, so we likely have a <strong>unit error<\/strong> here.)<\/li>\n<\/ul>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p>\u26a0\ufe0f <strong>Important Note<\/strong>: You wrote &#8220;827 \u03bcg&#8221; for pressure. Micrograms (\u03bcg) are <strong>not<\/strong> a unit of pressure. Could it have been meant to be <strong>827 mmHg<\/strong> or <strong>827 torr<\/strong>?<\/p>\n<\/blockquote>\n\n\n\n<p>Assuming you meant <strong>827 mmHg<\/strong> (which is a common pressure unit), let&#8217;s proceed by converting that to atm: P2=827&nbsp;mmHg760&nbsp;mmHg\/atm=1.088&nbsp;atmP_2 = \\frac{827 \\text{ mmHg}}{760 \\text{ mmHg\/atm}} = 1.088 \\text{ atm}<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Final Temperature T2=402T_2 = 402 K<\/li>\n\n\n\n<li>Final Volume V2=?V_2 = ?<\/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: Plug into the Combined Gas Law<\/h3>\n\n\n\n<p>P1V1T1=P2V2T2\u21d2V2=P1V1T2P2T1\\frac{P_1 V_1}{T_1} = \\frac{P_2 V_2}{T_2} \\Rightarrow V_2 = \\frac{P_1 V_1 T_2}{P_2 T_1} V2=(0.763\u2009atm)(71.3\u2009mL)(402\u2009K)(1.088\u2009atm)(310.45\u2009K)V_2 = \\frac{(0.763 \\, \\text{atm})(71.3 \\, \\text{mL})(402 \\, \\text{K})}{(1.088 \\, \\text{atm})(310.45 \\, \\text{K})}<\/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: Calculate<\/h3>\n\n\n\n<p>V2=(0.763)(71.3)(402)(1.088)(310.45)=21852.9582337.7236\u224864.7\u2009mLV_2 = \\frac{(0.763)(71.3)(402)}{(1.088)(310.45)} = \\frac{21852.9582}{337.7236} \\approx 64.7 \\, \\text{mL}<\/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>64.7&nbsp;mL\\boxed{64.7 \\text{ mL}}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83e\udde0 Explanation (300 words):<\/h3>\n\n\n\n<p>This problem involves understanding how a gas behaves under different conditions of pressure and temperature. The <strong>Combined Gas Law<\/strong> is a crucial formula derived from Boyle\u2019s Law, Charles\u2019s Law, and Gay-Lussac\u2019s Law. It allows you to calculate the change in volume of a gas when both the temperature and pressure are changing, provided the amount of gas remains constant.<\/p>\n\n\n\n<p>The initial conditions (volume, pressure, and temperature) describe a state of the gas, and we are asked to find the new volume after a change in pressure and temperature. Before using the formula, all units must be consistent: temperatures must be in <strong>Kelvin<\/strong> (not Celsius), and pressures must be in the <strong>same unit<\/strong> (here, we used atm).<\/p>\n\n\n\n<p>The pressure value &#8220;827 \u03bcg&#8221; initially given is likely a typo for <strong>827 mmHg<\/strong>, which is a valid pressure unit. Since the Combined Gas Law requires consistent units, mmHg was converted to atm using the conversion factor 1 atm = 760 mmHg.<\/p>\n\n\n\n<p>After conversion and substitution into the equation, we solved for the final volume. The result, <strong>64.7 mL<\/strong>, shows that the gas compresses slightly due to increased pressure and a higher temperature, which slightly offsets the pressure effect.<\/p>\n\n\n\n<p>This example shows the importance of careful unit conversion and a deep understanding of how gas particles behave under different thermodynamic conditions.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>A sample of argon gas has a volume of 71.3 mL at a pressure of 0.763 atm and a temperature of 37.3 C. What is the final volume of the gas, in mL, when the pressure and temperature are changed to 827 mug and 402 K The correct answer and explanation is: To solve this [&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-25853","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25853","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=25853"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25853\/revisions"}],"predecessor-version":[{"id":25858,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25853\/revisions\/25858"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=25853"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=25853"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=25853"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}