{"id":1528,"date":"2018-03-10T07:09:49","date_gmt":"2018-03-09T23:09:49","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1528"},"modified":"2023-08-19T02:00:20","modified_gmt":"2023-08-19T02:00:20","slug":"isothermal-thermodynamic-processes-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/isothermal-thermodynamic-processes-problems-and-solutions.htm","title":{"rendered":"Isothermal thermodynamic processes – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

30 Isothermal thermodynamic processes – problems and solutions<\/span><\/p>\n

1. PV diagram below shows an ideal gas<\/a> undergoes an isothermal process<\/a>. Calculate the work is done by the gas in the process AB.<\/span><\/span><\/p>\n

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

\"IsothermalWork<\/a> done by a gas is equal to the area under the PV curve<\/span><\/p>\n

AB = triangle area + rectangle area<\/span><\/p>\n

W = [\u00bd (8 x 105<\/sup>\u20134 x 105<\/sup>)(3-1)] + [4 x 105<\/sup> (3-1)] <\/span><\/p>\n

W = [\u00bd (4 x 105<\/sup>)(2)] + [4 x 105<\/sup> (2)]<\/span><\/p>\n

W = [4 x 105<\/sup>] + [8 x 105<\/sup>]<\/span><\/p>\n

W = 12 x 105<\/sup> Joule<\/span><\/p>\n

The work is done by the gas in the process AB = <\/span>12 x 105<\/sup> Joule<\/span><\/p>\n

2. Calculate the work is done by an ideal gas in the process ABC.<\/span><\/span><\/p>\n

\"IsothermalThe work is done by an ideal gas in the process ABC = <\/span>the area under the PV curve<\/span><\/span><\/p>\n

AB = triangle area + rectangle area<\/span><\/p>\n

W = [\u00bd(10×105<\/sup>\u20135×105<\/sup>)(30-10)]+[5×105<\/sup>(30-10)]<\/span><\/p>\n

W = [\u00bd (5 x 105<\/sup>)(20)] + [5 x 105<\/sup> (20)]<\/span><\/p>\n

W = [(5 x 105<\/sup>)(10)] + [100 x 105<\/sup>] <\/span><\/p>\n

W = [50 x 105<\/sup>] + [100 x 105<\/sup>] <\/span><\/p>\n

W = 150 x 105<\/sup> Joule<\/span><\/p>\n

W = 1.5 x 107<\/sup> Joule<\/span><\/p>\n

3. An ideal gas undergoing isothermal processes. What is an amount of heat<\/a> is added to the gas so the gas do work of 5000 Joule on the environment.<\/span><\/span><\/p>\n

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

Work (W) = 5000 Joule<\/span><\/p>\n

Wanted:<\/u> Heat is added to the gas (Q)<\/span><\/p>\n

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

An isothermal process is a<\/span> thermodynamic process that occurs at a constant temperature<\/a>. <\/span><\/p>\n

\u0394U = 3\/2 n R \u0394T<\/span><\/p>\n

\u0394U <\/i>= the change in internal energy, n = number of moles, R = universal gas constant, <\/i>\u0394T = The change in temperature.<\/i><\/span><\/p>\n

According to the above equation, if \u0394T = 0 then \u0394U = 0.<\/span><\/p>\n

The equation of the first law of thermodynamics :<\/u><\/span><\/p>\n

\u0394U = Q \u2013 W<\/span><\/p>\n

0 = Q \u2013 W<\/span><\/p>\n

Q = W<\/span><\/p>\n

Q = 5000 Joule.<\/span><\/p>\n

4. PV diagram for an ideal gas undergoing isothermal process shown in the figure below. Calculate the heat is added by a gas in process AB. <\/span><\/p>\n

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

Pressure 1 (P1<\/sub>) = 5 atm = 5 x 105<\/sup> Pa<\/span><\/p>\n

Pressure 2 (P2<\/sub>) = 10 atm = 10 x 105<\/sup> Pa<\/span><\/p>\n

Volume 1 (V1<\/sub>) = 2 m3<\/sup><\/span><\/p>\n

Volume 2 (V2<\/sub>) = 6 m3<\/sup><\/span><\/p>\n

Wanted<\/u> : Heat is added in process AB.<\/span><\/p>\n

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

Isothermal = constant temperature. According to the equation below, if \u0394T = 0 then \u0394U = 0. <\/span><\/p>\n

\u0394U = 3\/2 n R \u0394T<\/span><\/p>\n

\u0394U = 3\/2 n R (0)<\/span><\/p>\n

\u0394U = 0<\/span><\/p>\n

Apply to the first law of thermodynamics<\/a> :<\/span><\/p>\n

\u0394U = Q-W<\/span><\/p>\n

0 = Q-W<\/span><\/p>\n

Q=W<\/span><\/p>\n

The work is done by an ideal gas = <\/span>the area under the PV curve = triangle area + rectangle area<\/span><\/span><\/p>\n

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

W = \u00bd (10 x 105<\/sup> \u2013 5 x 105<\/sup>)(6-2) + (5 x 105<\/sup>)(6-2)<\/span><\/p>\n

W = \u00bd (5 x 105<\/sup>)(4) + (5 x 105<\/sup>)(4)<\/span><\/p>\n

W = \u00bd (20 x 105<\/sup>) + (20 x 105<\/sup>)<\/span><\/p>\n

W = (10 x 105<\/sup>) + (20 x 105<\/sup>)<\/span><\/p>\n

W = 30 x 105 <\/sup>Joule<\/span><\/p>\n

5. Calculate the work done on 1 mol of an ideal gas expanding isothermally from 2 L to 4 L at 300 K.<\/span>
\nSolution: \\( W = nRT \\ln\\left(\\frac{V_2}{V_1}\\right) = 1 \\times 8.314 \\times 300 \\times \\ln(2) \\approx 1724 \\, \\text{J} \\)<\/span><\/p>\n

6. Determine the heat transfer for the above problem.<\/span>
\nSolution: \\( Q = W = 1724 \\, \\text{J} \\)<\/span><\/p>\n

7. Calculate the change in internal energy for the above process.<\/span>
\nSolution: \\( \\Delta U = 0 \\, \\text{J} \\) (since the process is isothermal)<\/span><\/p>\n

8. For an isothermal compression of an ideal gas from 10 L to 5 L at 300 K, find the work done.<\/span>
\nSolution: \\( W = 8.314 \\times 300 \\times \\ln\\left(\\frac{5}{10}\\right) \\approx -862 \\, \\text{J} \\)<\/span><\/p>\n

9. Calculate the entropy change for the isothermal expansion in Problem 1.<\/span>
\nSolution: \\( \\Delta S = nR\\ln\\left(\\frac{V_2}{V_1}\\right) = 8.314 \\times \\ln(2) \\approx 5.76 \\, \\text{J\/K} \\)<\/span><\/p>\n

10. Find the heat transfer for an isothermal compression of 2 mol of an ideal gas from 4 L to 2 L at 200 K.<\/span>
\nSolution: \\( Q = 2 \\times 8.314 \\times 200 \\times \\ln\\left(\\frac{2}{4}\\right) \\approx -1152 \\, \\text{J} \\)<\/span><\/p>\n

11. Calculate the change in Gibbs free energy for an isothermal process.<\/span>
\nSolution: \\( \\Delta G = 0 \\) (For a reversible isothermal process in a closed system, \\( \\Delta G = 0 \\))<\/span><\/p>\n

12. Determine the entropy change for an isothermal compression of 3 mol of an ideal gas from 6 L to 3 L at 400 K.<\/span>
\nSolution: \\( \\Delta S = 3 \\times 8.314 \\times \\ln\\left(\\frac{3}{6}\\right) \\approx -17.29 \\, \\text{J\/K} \\)<\/span><\/p>\n

13. Find the work done by an ideal gas expanding isothermally from 3 L to 6 L at 250 K for 2 mol.<\/span>
\nSolution: \\( W = 2 \\times 8.314 \\times 250 \\times \\ln(2) \\approx 2874 \\, \\text{J} \\)<\/span><\/p>\n

14. Determine the heat transfer for the above process.<\/span>
\nSolution: \\( Q = W = 2874 \\, \\text{J} \\)<\/span><\/p>\n

15. Calculate the change in internal energy for the above isothermal expansion.<\/span>
\nSolution: \\( \\Delta U = 0 \\, \\text{J} \\)<\/span><\/p>\n

16. Calculate the entropy change for an isothermal expansion of 4 mol of an ideal gas from 5 L to 10 L at 500 K.<\/span>
\nSolution: \\( \\Delta S = 4 \\times 8.314 \\times \\ln(2) \\approx 23.03 \\, \\text{J\/K} \\)<\/span><\/p>\n

17. Determine the heat transfer for an isothermal compression of 1 mol of an ideal gas from 8 L to 4 L at 300 K.<\/span>
\nSolution: \\( Q = 8.314 \\times 300 \\times \\ln\\left(\\frac{4}{8}\\right) \\approx -862 \\, \\text{J} \\)<\/span><\/p>\n

18. Find the work done by 3 mol of an ideal gas expanding isothermally from 3 L to 9 L at 350 K.<\/span>
\nSolution: \\( W = 3 \\times 8.314 \\times 350 \\times \\ln(3) \\approx 5362 \\, \\text{J} \\)<\/span><\/p>\n

19. Calculate the entropy change for the above process.<\/span>
\nSolution: \\( \\Delta S = 3 \\times 8.314 \\times \\ln(3) \\approx 14.88 \\, \\text{J\/K} \\)<\/span><\/p>\n

20. Determine the heat transfer for an isothermal expansion of 2 mol of an ideal gas from 2 L to 8 L at 200 K.<\/span>
\nSolution: \\( Q = 2 \\times 8.314 \\times 200 \\times \\ln(4) \\approx 2304 \\, \\text{J} \\)<\/span><\/p>\n

21. Calculate the change in internal energy for an isothermal compression of 4 mol of an ideal gas from 10 L to 5 L at 500 K.<\/span>
\nSolution: \\( \\Delta U = 0 \\, \\text{J} \\)<\/span><\/p>\n

22. Determine the entropy change for an isothermal expansion of 5 mol of an ideal gas from 5 L to 15 L at 300 K.<\/span>
\nSolution: \\( \\Delta S = 5 \\times 8.314 \\times \\ln(3) \\approx 24.81 \\, \\text{J\/K} \\)<\/span><\/p>\n

23. Find the work done by 2 mol of an ideal gas compressing isothermally from 6 L to 2 L at 400 K.<\/span>
\nSolution: \\( W = 2 \\times 8.314 \\times 400 \\times \\ln\\left(\\frac{2}{6}\\right) \\approx -2874 \\, \\text{J} \\)<\/span><\/p>\n

24. Calculate the entropy change for an isothermal compression of 3 mol of an ideal gas from 9 L to 3 L at 250 K.<\/span>
\nSolution: \\( \\Delta S = 3 \\times 8.314 \\times \\ln\\left(\\frac{3}{9}\\right) \\approx -14.88 \\, \\text{J\/K} \\)<\/span><\/p>\n

25. Determine the heat transfer for an isothermal expansion of 1 mol of an ideal gas from 4 L to 12 L at 350 K.<\/span>
\nSolution: \\( Q = 8.314 \\times 350 \\times \\ln(3) \\approx 1791 \\, \\text{J} \\)<\/span><\/p>\n

26. Find the work done by 4 mol of an ideal gas expanding isothermally from 4 L to 8 L at 500 K.<\/span>
\nSolution: \\( W = 4 \\times 8.314 \\times 500 \\times \\ln(2) \\approx 5749 \\, \\text{J} \\)<\/span><\/p>\n

27. Determine the entropy change for an isothermal compression of 2\u00a0mol of an ideal gas from 10 L to 5 L at 200 K.<\/span>
\nSolution: \\( \\Delta S = 2 \\times 8.314 \\times \\ln\\left(\\frac{5}{10}\\right) \\approx -5.76 \\, \\text{J\/K} \\)<\/span><\/p>\n

24. Find the work done by 3 mol of an ideal gas compressing isothermally from 9 L to 3 L at 450 K.<\/span>
\nSolution: \\( W = 3 \\times 8.314 \\times 450 \\times \\ln\\left(\\frac{3}{9}\\right) \\approx -4310 \\, \\text{J} \\)<\/span><\/p>\n

25. Calculate the change in internal energy for an isothermal expansion of 5 mol of an ideal gas from 5 L to 10 L at 400 K.<\/span>
\nSolution: \\( \\Delta U = 0 \\, \\text{J} \\)<\/span><\/p>\n

26. Determine the entropy change for an isothermal expansion of 4 mol of an ideal gas from 6 L to 18 L at 300 K.<\/span>
\nSolution: \\( \\Delta S = 4 \\times 8.314 \\times \\ln(3) \\approx 19.85 \\, \\text{J\/K} \\)<\/span><\/p>\n

These problems cover various aspects of isothermal processes, including work done, heat transfer, changes in internal energy, and entropy change.<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

30 Isothermal thermodynamic processes – problems and solutions 1. PV diagram below shows an ideal gas undergoes an isothermal process. Calculate the work is done by the gas in the process AB. Solution Work done by a gas is equal to the area under the PV curve AB = triangle area + rectangle area W … 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":"Isothermal thermodynamic processes - 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-1528","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\/1528","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=1528"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1528\/revisions"}],"predecessor-version":[{"id":9024,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1528\/revisions\/9024"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1528"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1528"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1528"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}