{"id":1517,"date":"2018-03-09T15:15:53","date_gmt":"2018-03-09T07:15:53","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1517"},"modified":"2023-08-19T02:48:45","modified_gmt":"2023-08-19T02:48:45","slug":"application-of-the-first-law-of-thermodynamics-in-some-thermodynamics-processes-isobaric-isothermal-isochoric","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/application-of-the-first-law-of-thermodynamics-in-some-thermodynamics-processes-isobaric-isothermal-isochoric.htm","title":{"rendered":"Application of the first law of thermodynamics in some thermodynamic processes (Isobaric Isothermal Isochoric)","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

30 Application of the first law of thermodynamics in some thermodynamic processes (Isobaric Isothermal Isochoric)<\/strong><\/span><\/p>\n

1. The graph below shows the thermodynamic<\/a> cycle experienced by a gas. The <\/span>work<\/a> done by <\/span>gas on the process ABCD is \u2026<\/span><\/span><\/p>\n

\"ApplicationKnown :<\/u><\/span><\/p>\n

Pressure<\/a> 1 (P1<\/sub>) = 2 x 105<\/sup> Pa<\/span><\/p>\n

Pressure 2 (P2<\/sub>) = 4 x 105<\/sup> Pa<\/span><\/p>\n

Volume 1 (V1<\/sub>) = 1 m3<\/sup><\/span><\/p>\n

Volume 2 (V2<\/sub>) = 3 m3<\/sup><\/span><\/p>\n

Wanted: Work done by gas on process ABCD (W).<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Work done by a gas is equal to the area of ABCD. <\/span><\/p>\n

W = (P2<\/sub> \u2013 P1<\/sub>)(V2<\/sub> \u2013 V1<\/sub>) <\/span><\/p>\n

W = (4 x 105<\/sup> \u2013 2 x 105<\/sup>)(3 \u2013 1)<\/span><\/p>\n

W = (2 x 105<\/sup>)(2) <\/span><\/p>\n

W = 4 x 105<\/sup> Joule<\/span><\/p>\n

2. The work done by gas on process ABC is…<\/span><\/p>\n

\"ApplicationKnown :<\/u><\/span><\/p>\n

Pressure 1 (P1<\/sub>) = 3 x 105<\/sup> Pa<\/span><\/p>\n

Pressure 2 (P2<\/sub>) = 6 x 105<\/sup> Pa<\/span><\/p>\n

Volume 1 (V1<\/sub>) = 20 cm3<\/sup> = 20 x 10-6<\/sup> m3<\/sup><\/span><\/p>\n

Volume 2 (V2<\/sub>) = 60 cm3<\/sup> = 60 x 10-6<\/sup> m3<\/sup><\/span><\/p>\n

Wanted <\/u>: The work done by gas on process ABC <\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Work done by gas = area of ABC. <\/span><\/p>\n

W = \u00bd (P2<\/sub> \u2013 P1<\/sub>)(V2<\/sub> \u2013 V1<\/sub>) <\/span><\/p>\n

W = 1\/2 (6 x 105<\/sup> \u2013 3 x 105<\/sup>)(60 x 10-6<\/sup> \u2013 20 x 10-6<\/sup>)<\/span><\/p>\n

W = 1\/2 (3 x 105<\/sup>)(40 x 10-6<\/sup>) <\/span><\/p>\n

W = 1\/2 (120 x 10-1<\/sup> Joule)<\/span><\/p>\n

W = 1\/2 (12 Joule)<\/span><\/p>\n

W = 6 Joule<\/span><\/p>\n

3. PV diagram for a gas in a closed container shown in figure below. The work done by the gas is shown in which process.<\/span><\/span><\/p>\n

\"ApplicationSolution<\/span><\/p>\n

P<\/span>rocess AB and process DC <\/span>are <\/span>isobaric processes <\/span>(the pressure is kept constant)<\/span>. <\/span>P<\/span>rocess AD <\/span>and <\/span>process BC <\/span>are i<\/a><\/span>so<\/span>c<\/span>horic processes <\/span>(volume does not change)<\/span>.<\/span><\/span><\/p>\n

The work was done by gas when the gas expands (process DC).<\/span><\/span><\/p>\n

Here are 9 problems and solutions related to the Isobaric Process (Constant Pressure):<\/span><\/p>\n

Problem 1:<\/span>
\nA 3.0 m\u00b3 gas at constant pressure is compressed to 2.0 m\u00b3. If the initial temperature is 300 K, what is the final temperature?<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for an isobaric process: \\(\\frac{V_1}{T_1} = \\frac{V_2}{T_2}\\), we can solve for the final temperature:<\/span><\/p>\n

\\[<\/span>
\n\\frac{3.0}{300} = \\frac{2.0}{T_2} \\implies T_2 = \\frac{2.0 \\times 300}{3.0} = 200\\, \\text{K}<\/span>
\n\\]<\/span><\/p>\n

Problem 2:<\/span>
\nCalculate the work done by 2 moles of gas expanding isobarically from 1 liter to 3 liters at 2 atm.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for work done in an isobaric process: \\( W = P \\Delta V \\):<\/span><\/p>\n

\\[<\/span>
\nW = 2\\, \\text{atm} \\times (3 – 1)\\, \\text{liters} = 4\\, \\text{atm}\\cdot\\text{liters}<\/span>
\n\\]<\/span><\/p>\n

Problem 3:<\/span>
\nWhat is the change in internal energy when 4 moles of an ideal diatomic gas are heated at constant pressure from 200 K to 300 K?<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for the change in internal energy at constant pressure for a diatomic gas: \\( \\Delta U = nC_pdT \\), where \\( C_p = \\frac{7}{2}R \\):<\/span><\/p>\n

\\[<\/span>
\n\\Delta U = 4 \\times \\frac{7}{2} \\times 8.314 \\times (300 – 200) \\approx 6990\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 4:<\/span>
\nCalculate the heat added during an isobaric process when 2 moles of a monatomic gas expand from 5 liters to 10 liters at 400 K.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for heat transfer in an isobaric process: \\( Q = nC_pdT \\), where \\( C_p = \\frac{5}{2}R \\), and \\( \\frac{V_2}{T_2} = \\frac{V_1}{T_1} \\):<\/span><\/p>\n

\\[<\/span>
\nQ = 2 \\times \\frac{5}{2} \\times 8.314 \\times \\left(400 \\times \\frac{10}{5} – 400\\right) = 3316\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 5:<\/span>
\nDetermine the change in enthalpy of 3 moles of a triatomic gas that is expanded from 1 liter to 2 liters at a temperature of 200 K.<\/span><\/p>\n

Solution:<\/span>
\nUsing the formula for the change in enthalpy for an isobaric process: \\( \\Delta H = nC_pdT \\), where \\( C_p = \\frac{f}{2}R \\) and \\( f = 6 \\):<\/span><\/p>\n

\\[<\/span>
\n\\Delta H = 3 \\times \\frac{6}{2} \\times 8.314 \\times \\left(200 \\times \\frac{2}{1} – 200\\right) = 4989\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 6:<\/span>
\nCalculate the final volume of a 2 moles of a gas at 300 K and 5 liters initially when it is heated to 500 K at constant pressure.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for an isobaric process:<\/span><\/p>\n

\\[<\/span>
\n\\frac{V_1}{T_1} = \\frac{V_2}{T_2} \\implies V_2 = \\frac{V_1 \\times T_2}{T_1} = \\frac{5 \\times 500}{300} \\approx 8.33\\, \\text{liters}<\/span>
\n\\]<\/span><\/p>\n

Problem 7:<\/span>
\nFind the work done by a monatomic gas when it is compressed from 6 liters to 3 liters at 2 atm.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for work done in an isobaric process:<\/span><\/p>\n

\\[<\/span>
\nW = P \\Delta V = 2 \\times (3 – 6) = -6\\, \\text{atm}\\cdot\\text{liters}<\/span>
\n\\]<\/span><\/p>\n

Problem 8:<\/span>
\nA diatomic gas undergoes an isobaric expansion from 3 liters to 6 liters at 300 K. Find the change in entropy.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for the change in entropy in an isobaric process for a diatomic gas, where \\( C_p = \\frac{7}{2}R \\):<\/span><\/p>\n

\\[<\/span>
\n\\Delta S = nC_p\\ln\\frac{T_2}{T_1} + nR\\ln\\frac{V_2}{V_1} = 2 \\times \\frac{7}{2} \\times 8.314 \\times \\ln\\frac{300 \\times \\frac{6}{3}}{300} + 2 \\times 8.314 \\times \\ln\\frac{6}{3} \\approx 34.76\\, \\text{J\/K}<\/span>
\n\\]<\/span><\/p>\n

Problem 9:<\/span>
\nCalculate the heat transfer during an isobaric compression of 5 moles of a triatomic gas from 10 liters to 5 liters at 500 K.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for heat transfer in an isobaric process, where \\( C_p = \\frac{f}{2}R \\) and \\( f = 6 \\):<\/span><\/p>\n

\\[<\/span>
\nQ = 5 \\times \\frac{6}{2} \\times 8.314 \\times \\left(500 \\times \\frac{5}{10} – 500\\right) = -12473\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

These problems should give a comprehensive overview of various concepts related to the isobaric process.<\/span><\/p>\n

Here are 9 problems and solutions related to the Isothermal Process (Constant Temperature):<\/span><\/p>\n

\u00a0Problem 1:<\/span>
\nProblem: Calculate the work done by 2 moles of an ideal gas expanding isothermally from 1 liter to 2 liters at 300 K.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for work done in an isothermal process for an ideal gas: <\/span>
\n\\[ W = nRT\\ln\\frac{V_2}{V_1} \\]<\/span><\/p>\n

\\[<\/span>
\nW = 2 \\times 8.314 \\times 300 \\times \\ln\\frac{2}{1} \\approx 3454\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 2:<\/span>
\nProblem: Find the heat transferred when 3 moles of a gas are compressed isothermally from 4 liters to 2 liters at 400 K.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for heat transfer in an isothermal process: \\( Q = W \\):<\/span><\/p>\n

\\[<\/span>
\nQ = 3 \\times 8.314 \\times 400 \\times \\ln\\frac{2}{4} \\approx -3462\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 3:<\/span>
\nDetermine the change in entropy when 4 moles of a gas undergo an isothermal expansion from 1 liter to 5 liters at 300 K.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for the change in entropy in an isothermal process:<\/span><\/p>\n

\\[<\/span>
\n\\Delta S = nR\\ln\\frac{V_2}{V_1} = 4 \\times 8.314 \\times \\ln\\frac{5}{1} \\approx 46.15\\, \\text{J\/K}<\/span>
\n\\]<\/span><\/p>\n

Problem 4:<\/span>
\nCalculate the work done when 1 mole of an ideal gas is compressed isothermally from 6 liters to 3 liters at 200 K.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for work done in an isothermal process:<\/span><\/p>\n

\\[<\/span>
\nW = 1 \\times 8.314 \\times 200 \\times \\ln\\frac{3}{6} \\approx -575\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 5:<\/span>
\nDetermine the change in internal energy for 3 moles of gas during an isothermal expansion from 2 liters to 6 liters at constant temperature.<\/span><\/p>\n

Solution:<\/span>
\nFor an isothermal process of an ideal gas, the change in internal energy is zero:<\/span><\/p>\n

\\[<\/span>
\n\\Delta U = 0\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 6:<\/span>
\nCalculate the heat transfer when 5 moles of a gas are expanded isothermally from 3 liters to 6 liters at 250 K.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for heat transfer in an isothermal process:<\/span><\/p>\n

\\[<\/span>
\nQ = 5 \\times 8.314 \\times 250 \\times \\ln\\frac{6}{3} \\approx 2879\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 7:<\/span>
\nWhat is the work done by 2 moles of a gas during an isothermal compression from 4 liters to 2 liters at 150 K?<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for work done in an isothermal process:<\/span><\/p>\n

\\[<\/span>
\nW = 2 \\times 8.314 \\times 150 \\times \\ln\\frac{2}{4} \\approx -1151\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 8:<\/span>
\nFind the change in entropy when 3 moles of a gas are compressed isothermally from 5 liters to 1 liter at 500 K.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for the change in entropy in an isothermal process:<\/span><\/p>\n

\\[<\/span>
\n\\Delta S = 3 \\times 8.314 \\times \\ln\\frac{1}{5} \\approx -34.77\\, \\text{J\/K}<\/span>
\n\\]<\/span><\/p>\n

Problem 9:<\/span>
\nDetermine the work done by 4 moles of an ideal gas expanding isothermally from 2 liters to 8 liters at 100 K.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for work done in an isothermal process:<\/span><\/p>\n

\\[<\/span>
\nW = 4 \\times 8.314 \\times 100 \\times \\ln\\frac{8}{2} \\approx 2304\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

These problems cover various concepts related to the isothermal process, such as work done, heat transfer, and change in entropy.<\/span><\/p>\n

Here are 9 problems and solutions related to the Isochoric Process (Constant Volume):<\/span><\/p>\n

Problem 1:<\/span>
\nCalculate the change in internal energy for 3 moles of a monatomic ideal gas when the temperature is increased from 200 K to 400 K at constant volume.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for change in internal energy at constant volume: \\( \\Delta U = nC_v\\Delta T \\), where \\( C_v = \\frac{3}{2}R \\):<\/span><\/p>\n

\\[<\/span>
\n\\Delta U = 3 \\times \\frac{3}{2} \\times 8.314 \\times (400 – 200) \\approx 3741\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 2:<\/span>
\nDetermine the heat transfer for 2 moles of a diatomic gas when the temperature is decreased from 300 K to 200 K at constant volume.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for heat transfer at constant volume: \\( Q = \\Delta U = nC_v\\Delta T \\), where \\( C_v = \\frac{5}{2}R \\):<\/span><\/p>\n

\\[<\/span>
\nQ = 2 \\times \\frac{5}{2} \\times 8.314 \\times (200 – 300) \\approx -4157\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 3:<\/span>
\nCalculate the change in entropy for 4 moles of a triatomic gas when the temperature increases from 100 K to 300 K at constant volume.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for the change in entropy at constant volume, where \\( C_v = \\frac{f}{2}R \\) and \\( f = 6 \\):<\/span><\/p>\n

\\[<\/span>
\n\\Delta S = nC_v\\ln\\frac{T_2}{T_1} = 4 \\times \\frac{6}{2} \\times 8.314 \\times \\ln\\frac{300}{100} \\approx 115.36\\, \\text{J\/K}<\/span>
\n\\]<\/span><\/p>\n

Problem 4:<\/span>
\nWhat is the work done by a gas during an isochoric process?<\/span><\/p>\n

Solution:<\/span>
\nSince the volume is constant during an isochoric process, there is no work done:<\/span><\/p>\n

\\[<\/span>
\nW = 0\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 5:<\/span>
\nDetermine the heat transfer for 5 moles of a monatomic ideal gas when it is cooled from 500 K to 300 K at constant volume.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for heat transfer at constant volume, where \\( C_v = \\frac{3}{2}R \\):<\/span><\/p>\n

\\[<\/span>
\nQ = 5 \\times \\frac{3}{2} \\times 8.314 \\times (300 – 500) \\approx -6232\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 6:<\/span>
\nCalculate the change in internal energy for 1 mole of a diatomic gas when the temperature is increased from 150 K to 250 K at constant volume.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for change in internal energy at constant volume, where \\( C_v = \\frac{5}{2}R \\):<\/span><\/p>\n

\\[<\/span>
\n\\Delta U = 1 \\times \\frac{5}{2} \\times 8.314 \\times (250 – 150) \\approx 2079\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 7:<\/span>
\nDetermine the change in entropy for 3 moles of a monatomic gas when the temperature decreases from 600 K to 300 K at constant volume.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for the change in entropy at constant volume, where \\( C_v = \\frac{3}{2}R \\):<\/span><\/p>\n

\\[<\/span>
\n\\Delta S = 3 \\times \\frac{3}{2} \\times 8.314 \\times \\ln\\frac{300}{600} \\approx -34.59\\, \\text{J\/K}<\/span>
\n\\]<\/span><\/p>\n

Problem 8:<\/span>
\nWhat is the heat transfer for 2 moles of a triatomic gas when the temperature is increased from 200 K to 400 K at constant volume?<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for heat transfer at constant volume, where \\( C_v = \\frac{6}{2}R \\):<\/span><\/p>\n

\\[<\/span>
\nQ = 2 \\times \\frac{6}{2} \\times 8.314 \\times (400 – 200) \\approx 4986\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

Problem 9:<\/span>
\nCalculate the change in internal energy for 4 moles of a monatomic ideal gas when it is heated from 250 K to 350 K at constant volume.<\/span><\/p>\n

Solution:<\/span>
\nUsing the equation for change in internal energy at constant volume, where \\( C_v = \\frac{3}{2}R \\):<\/span><\/p>\n

\\[<\/span>
\n\\Delta U = 4 \\times \\frac{3}{2} \\times 8.314 \\times (350 – 250) \\approx 4986\\, \\text{J}<\/span>
\n\\]<\/span><\/p>\n

These problems cover various aspects of the isochoric process, such as heat transfer, changes in internal energy, and entropy, for different types of gases.<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

30 Application of the first law of thermodynamics in some thermodynamic processes (Isobaric Isothermal Isochoric) 1. The graph below shows the thermodynamic cycle experienced by a gas. The work done by gas on the process ABCD is \u2026 Known : Pressure 1 (P1) = 2 x 105 Pa Pressure 2 (P2) = 4 x 105 … 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":"Application of the first law of thermodynamics in some thermodynamic processes (isobaric isothermal isochoric)","_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-1517","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\/1517","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=1517"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1517\/revisions"}],"predecessor-version":[{"id":9025,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1517\/revisions\/9025"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1517"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1517"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1517"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}