{"id":2330,"date":"2018-05-01T10:18:49","date_gmt":"2018-05-01T02:18:49","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=2330"},"modified":"2023-08-08T12:32:32","modified_gmt":"2023-08-08T12:32:32","slug":"work-done-in-thermodynamics-process-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/work-done-in-thermodynamics-process-problems-and-solutions.htm","title":{"rendered":"Work done in thermodynamics process \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"<p style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Work done in thermodynamics process \u2013 problems and solutions<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">1. An <a href=\"https:\/\/gurumuda.net\/physics\/ideal-gas-law-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">ideal gas<\/a> is compressed from state 1 to state 2 as shown in the graph below. Determine the <a href=\"https:\/\/gurumuda.net\/physics\/work-done-by-force-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">work<\/a> done by the gas.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Solution<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">1 liter = 0.001 m<sup>3 <\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Known :<img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-2331\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/05\/Work-done-in-thermodynamics-process-\u2013-problems-and-solutions-1.png\" alt=\"Work done in thermodynamics process \u2013 problems and solutions 1\" width=\"208\" height=\"145\" \/><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><a href=\"https:\/\/gurumuda.net\/physics\/pressure-of-fluids-problems-and-solutions.htm\" target=\"_blank\" rel=\"noopener\">Pressure<\/a> 1 (P<sub>1<\/sub>) = 10 kPa = 10,000 Pa = 10,000 N\/m<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Pressure 2 (P<sub>2<\/sub>) = 20 kPa = 20,000 Pa = 20,000 N\/m<sup>2<\/sup><!--more--><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Volume 1 (V<sub>1<\/sub>) = 4 liters = 4 (0.001 m<sup>3<\/sup>) = 0.004 m<sup>3<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Volume 2 (V<sub>2<\/sub>) = 10 liters = 10 (0.001 m<sup>3<\/sup>) = 0.010 m<sup>3<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted :<\/u> The work done by the gas (W)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">W = P (V<sub>2<\/sub> \u2013 V<sub>1<\/sub>) + 1\/2 (P<sub>2<\/sub> \u2013 P<sub>1<\/sub>)(V<sub>2<\/sub> \u2013 V<sub>1<\/sub>) <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">W = 10,000 (0.010-0.004) + 1\/2 (20,000-10,000)(0.010-0.004) <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">W = 10,000 (0.006) + 1\/2 (10,000)(0.006) <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">W = 10 (6) + 1\/2 (10)(6) <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">W = 60 + 1\/2 (60) <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">W = 60 + 30<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">W = 90 Joule<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">2. Determine the work done by the gas in process a-b-c-d-a, as shown in graph below.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">Known :<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Pressure 1 (P<sub>1<\/sub>) = 1 x 10<sup>5<\/sup> Pa = 1 x 10<sup>5<\/sup> N\/m<sup>2<img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-2332\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/05\/Work-done-in-thermodynamics-process-\u2013-problems-and-solutions-2.png\" alt=\"Work done in thermodynamics process \u2013 problems and solutions 2\" width=\"176\" height=\"159\" \/><\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Pressure 2 (P<sub>2<\/sub>) = 3 x 10<sup>5<\/sup> Pa = 3 x 10<sup>5<\/sup> N\/m<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Volume 1 (V<sub>1<\/sub>) = 2 m<sup>3<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Volume 2 (V<sub>2<\/sub>) = 4 m<sup>3<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted :<\/u> Work (W)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The work done by the gas = area of rectangular a-b-c-d. <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">W = (P<sub>2<\/sub> \u2013 P<sub>1<\/sub>)(V<sub>2<\/sub> \u2013 V<sub>1<\/sub>) <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">W = (3 x 10<sup>5<\/sup> \u2013 1 x 10<sup>5<\/sup>)(4 \u2013 2)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">W = (2 x 10<sup>5<\/sup>)(2) <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">W = 4 x 10<sup>5<\/sup> Joule<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">W = 400,000 Joule<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">W = 400 kJ<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">3. Determine the work done by the gas in process A-B-C-A.<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Known :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Pressure 1 (P<sub>1<\/sub>) = 1 N\/m<sup>2<img loading=\"lazy\" decoding=\"async\" class=\"alignright size-full wp-image-2333\" src=\"https:\/\/gurumuda.net\/physics\/wp-content\/uploads\/2018\/05\/Work-done-in-thermodynamics-process-\u2013-problems-and-solutions-3.png\" alt=\"Work done in thermodynamics process \u2013 problems and solutions 3\" width=\"186\" height=\"162\" \/><\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Pressure 2 (P<sub>2<\/sub>) = 7 N\/m<sup>2<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Volume 1 (V<sub>1<\/sub>) = 1 m<sup>3<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">Volume 2 (V<sub>2<\/sub>) = 5 m<sup>3<\/sup><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><u>Wanted :<\/u> Work (W)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\"><u>Solution :<\/u><\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">The work done by the gas = area of triangle A-B-C<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">W = \u00bd (P<sub>2<\/sub> \u2013 P<sub>1<\/sub>)(V<sub>2<\/sub> \u2013 V<sub>1<\/sub>) <\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">W = \u00bd (7-1)(5-1)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\">W = \u00bd (6)(4)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">W = \u00bd (24)<\/span><\/p>\n<p class=\"western\" style=\"text-align: justify;\" align=\"justify\"><span style=\"font-family: 'times new roman', times, serif; font-size: 12pt;\">W = 12 Joule<\/span><\/p>\n<ol style=\"text-align: justify;\">\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is the definition of work in thermodynamics?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> In thermodynamics, work is defined as an energy transfer to or from a system that is not due to a temperature difference between the system and its surroundings.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How is work related to the pressure-volume (P-V) diagram of a thermodynamic process?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> In a P-V diagram, the work done by or on a system during a process is represented by the area under the curve of the process. For processes involving gases, the work is often given by the integral of P dV, where P is the pressure and dV is the differential change in volume.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is the work done during an isochoric process?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> An isochoric process occurs at constant volume. Since there&#8217;s no volume change, the work done during an isochoric process is zero.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does the work done in an isobaric (constant pressure) expansion differ from that in an isothermal (constant temperature) expansion of an ideal gas?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> In an isobaric expansion, work is calculated as <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">W<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">P<\/span><span class=\"mbin\">\u00d7<\/span><\/span><span class=\"base\"><span class=\"mord\">\u0394<\/span><span class=\"mord mathnormal\">V<\/span><\/span><\/span><\/span><\/span>, where P is the constant pressure and <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\">\u0394<\/span><span class=\"mord mathnormal\">V<\/span><\/span><\/span><\/span><\/span> is the change in volume. For an isothermal expansion of an ideal gas, work can be calculated using the equation <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord mathnormal\">W<\/span><span class=\"mrel\">=<\/span><\/span><span class=\"base\"><span class=\"mord mathnormal\">n<\/span><span class=\"mord mathnormal\">RT<\/span><span class=\"mop\">ln<\/span><span class=\"minner\"><span class=\"mopen delimcenter\"><span class=\"delimsizing size2\">(<\/span><\/span><span class=\"mord\"><span class=\"mfrac\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mtight\"><span class=\"mord mathnormal mtight\">V<\/span><span class=\"msupsub\"><span class=\"sizing reset-size3 size1 mtight\"><span class=\"mord mathnormal mtight\">f\/Vi<span class=\"vlist-s\">\u200b<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><span class=\"mclose delimcenter\"><span class=\"delimsizing size2\">)<\/span><\/span><\/span><\/span><\/span><\/span><\/span>, where n is the number of moles, R is the universal gas constant, and <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mord mathnormal\">V<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">f<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> and <span class=\"math math-inline\"><span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"mord\"><span class=\"mord mathnormal\">V<\/span><span class=\"msupsub\"><span class=\"vlist-t vlist-t2\"><span class=\"vlist-r\"><span class=\"vlist\"><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mord mathnormal mtight\">i<\/span><\/span><\/span><span class=\"vlist-s\">\u200b<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span> are the final and initial volumes, respectively.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What is the significance of the first law of thermodynamics regarding work and energy?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> The first law of thermodynamics states that the change in the internal energy of a system is equal to the heat added to the system minus the work done by the system on its surroundings. It represents the conservation of energy principle.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>For a cyclic process on a P-V diagram, what does the enclosed area represent?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> For a cyclic process on a P-V diagram, the enclosed area represents the net work done by or on the system during one cycle.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does the work done in an adiabatic process differ from an isothermal process for an ideal gas?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> In an adiabatic process, there is no heat exchange with the surroundings, so the work done results in a change in the internal energy of the gas. For an isothermal process, the temperature remains constant, and any work done is balanced by an equal amount of heat transfer, so there&#8217;s no change in internal energy.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Why can the work done in an irreversible process be different from a reversible process even if the initial and final states are the same?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> In a reversible process, each intermediate state between the initial and final states is in equilibrium. In an irreversible process, this isn&#8217;t the case. The paths taken on a P-V diagram, and thus the work done, can differ significantly between the two types of processes, even if they start and end at the same points.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>How does the presence of non-ideal (real) gas behavior affect work calculations in thermodynamic processes?<\/strong><\/span>\n<ul>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> For real gases, deviations from ideal behavior can lead to differences in work calculations, especially at high pressures or low temperatures. The Van der Waals equation or other real gas models might need to be used to account for these deviations.<\/span><\/li>\n<\/ul>\n<\/li>\n<li><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>What does it mean when the work done on a system is negative?<\/strong><\/span><\/li>\n<\/ol>\n<ul>\n<li style=\"text-align: justify;\"><span style=\"font-size: 12pt; font-family: 'times new roman', times, serif;\"><strong>Answer:<\/strong> When the work done on a system is negative, it means the system is doing work on its surroundings. This often corresponds to an expansion of a gas against external pressure.<\/span><\/li>\n<\/ul>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"<p>Work done in thermodynamics process \u2013 problems and solutions 1. An ideal gas is compressed from state 1 to state 2 as shown in the graph below. Determine the work done by the gas. Solution 1 liter = 0.001 m3 Known : Pressure 1 (P1) = 10 kPa = 10,000 Pa = 10,000 N\/m2 Pressure &#8230; <a title=\"Work done in thermodynamics process \u2013 problems and solutions\" class=\"read-more\" href=\"https:\/\/gurumuda.net\/physics\/work-done-in-thermodynamics-process-problems-and-solutions.htm\" aria-label=\"Read more about Work done in thermodynamics process \u2013 problems and solutions\">Read more<\/a><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"author":1,"featured_media":0,"comment_status":"closed","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":"Work done in thermodynamics process \u2013 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-2330","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\/2330","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=2330"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2330\/revisions"}],"predecessor-version":[{"id":8595,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/2330\/revisions\/8595"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=2330"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=2330"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=2330"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}