{"id":1569,"date":"2018-03-10T14:18:09","date_gmt":"2018-03-10T06:18:09","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1569"},"modified":"2023-08-19T01:36:08","modified_gmt":"2023-08-19T01:36:08","slug":"isobaric-thermodynamics-processes-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/isobaric-thermodynamics-processes-problems-and-solutions.htm","title":{"rendered":"Isobaric thermodynamics processes – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

30 Isobaric thermodynamics processes – problems and solutions<\/span><\/p>\n

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

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

Pressure<\/a> (P) = 5 x 105<\/sup> N\/m2<\/sup><\/span><\/p>\n

Initial volume (V1<\/sub>) = 2 m3<\/sup><\/span><\/p>\n

Final volume (V2<\/sub>) = 6 m3<\/sup><\/span><\/p>\n

Wanted :<\/u> Work (W)<\/span><\/p>\n

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

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

W = (5 x 105<\/sup>)(6 – 2) = (5 x 105<\/sup>) (4)<\/span><\/p>\n

W = 20 x 105 <\/sup>= 2 x 106<\/sup> Joule <\/span><\/p>\n

2. What is difference of the work is done by the gas in process AB and process CD… <\/span><\/p>\n

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

Isobaric process <\/b>AB<\/b> :<\/span><\/p>\n

Pressure (P) = 6 atm = 6 x 105<\/sup> N\/m2<\/sup><\/span><\/p>\n

Initial volume (V1<\/sub>) = 1 liter = 1 dm3<\/sup> = 1 x 10-3<\/sup> m3<\/sup><\/span><\/p>\n

Final volume (V2<\/sub>) = 3 liters = 3 dm3<\/sup> = 3 x 10-3<\/sup> m3 <\/sup><\/span><\/p>\n

Isobaric process CD <\/b>:<\/span><\/p>\n

Pressure (P) = 4 atm = 4 x 105<\/sup> N\/m2<\/sup><\/span><\/p>\n

Initial volume (V1<\/sub>) = 2 liters = 2 dm3<\/sup> = 2 x 10-3<\/sup> m3<\/sup><\/span><\/p>\n

Final volume (V2<\/sub>) = 5 liters = 5 dm3<\/sup> = 5 x 10-3<\/sup> m3<\/sup><\/span><\/p>\n

Wanted <\/u>:<\/u> Difference of the work is done by the gas in process AB and CD.<\/span><\/p>\n

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

Work is done by the gas in process AB :<\/span><\/p>\n

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

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

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

W = 12 x 102 <\/sup>= 1200 Joule<\/span><\/p>\n

Work is done by the gas in process CD :<\/span><\/p>\n

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

W = (4 x 105<\/sup>)(5 x 10-3 <\/sup>– 2 x 10-3<\/sup>) <\/span><\/p>\n

W = (4 x 105<\/sup>)(3 x 10-3<\/sup>)<\/span><\/p>\n

W = 12 x 102 <\/sup>= 1200 Joule<\/span><\/p>\n

Difference of the work is done by the gas in process AB and CD = 1200 \u2013 1200 = 0.<\/span><\/p>\n

3. Work is done by the gas in process ABC is….<\/span><\/p>\n

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

Pressure 1 (P1<\/sub>) = 6 x 105<\/sup> Pa = 6 x 105<\/sup> N\/m2<\/sup><\/span><\/p>\n

Pressure 2 (P2<\/sub>) = 3 x 105 <\/sup>Pa = 3 x 105<\/sup> N\/m2<\/sup><\/span><\/p>\n

Volume 1 (V1<\/sub>) = 2 cm3<\/sup> = 2 x 10-6<\/sup> m3<\/sup><\/span><\/p>\n

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

Wanted<\/u> : Work is done in process ABC.<\/span><\/p>\n

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

In process AB, the volume is kept constant so that no work is done by the gas. <\/span><\/p>\n

Work was done by the gas in the process BC.<\/span><\/p>\n

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

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

W = (3 x 105<\/sup>)(4 x 10-6<\/sup>)<\/span><\/p>\n

W = 12 x 10-1<\/sup><\/span><\/p>\n

W = 1.2 Joule<\/span><\/p>\n

Work is done in the process ABC = work is done in the process AB = 1.2 Joule.<\/span><\/p>\n

4. Determine the change in internal energy for 2 moles of an ideal gas undergoing an isobaric expansion at 300 K, where \\(\\Delta V = 1\\ \\text{m}^3\\).<\/span>
\nSolution: \\(\\Delta U = nC_v\\Delta T\\), using \\(C_v = \\frac{R}{\\gamma-1}\\) (for monatomic ideal gas, \\(\\gamma = \\frac{5}{3}\\)) and \\(\\Delta T = \\frac{P\\Delta V}{nR}\\), \\(\\Delta U = \\frac{2\\cdot 300 \\cdot 1}{\\frac{5}{3}-1} \\approx 1800\\ \\text{J}\\).<\/span><\/p>\n

5. Calculate the heat transfer in an isobaric process where 1 mole of a diatomic ideal gas expands, \\(C_p = \\frac{7}{2}R\\), and \\(\\Delta T = 50\\ \\text{K}\\).<\/span>
\nSolution: \\(Q = nC_p\\Delta T = \\frac{7}{2} \\cdot 50 \\cdot R \\approx 1750\\ \\text{J}\\) (using \\(R = 8.314\\ \\text{J\/(mol\u00b7K)}\\)).<\/span><\/p>\n

6. Find the work done by a system undergoing an isobaric expansion, \\(P = 3\\ \\text{atm}\\), \\(\\Delta V = 4\\ \\text{L}\\).<\/span>
\nSolution: \\(W = P\\Delta V = 3 \\times 4 = 12\\ \\text{L\u00b7atm}\\).<\/span><\/p>\n

7. Determine the change in entropy for an isobaric process where 2 moles of an ideal gas change temperature by 20 K. Use \\(C_p = \\frac{5}{2}R\\).<\/span>
\nSolution: \\(\\Delta S = nC_p\\ln\\frac{T_2}{T_1} = 2 \\cdot \\frac{5}{2}R \\cdot \\ln\\frac{T_1+20}{T_1}\\).<\/span><\/p>\n

8. Calculate the heat transfer for an isobaric compression of a monatomic ideal gas, \\(C_p = \\frac{5}{2}R\\), \\(\\Delta T = -10\\ \\text{K}\\).<\/span>
\nSolution: \\(Q = nC_p\\Delta T = \\frac{5}{2} \\cdot (-10) \\cdot R \\approx -415\\ \\text{J}\\).<\/span><\/p>\n

9. Find the work done on the system in an isobaric process with \\(P = 5\\ \\text{bar}\\), \\(\\Delta V = -3\\ \\text{m}^3\\).<\/span>
\nSolution: \\(W = P\\Delta V = 5 \\times (-3) = -15\\ \\text{bar\u00b7m}^3\\).<\/span><\/p>\n

10. Determine the change in internal energy for an isobaric process where \\(n = 3\\ \\text{mol}\\), \\(C_v = 3R\\), \\(\\Delta T = 25\\ \\text{K}\\).<\/span>
\nSolution: \\(\\Delta U = nC_v\\Delta T = 3 \\cdot 3R \\cdot 25 \\approx 1883\\ \\text{J}\\).<\/span><\/p>\n

11. Calculate the entropy change in an isobaric process for a diatomic ideal gas, \\(n = 1\\ \\text{mol}\\), \\(\\Delta T = 40\\ \\text{K}\\), \\(T_1 = 300\\ \\text{K}\\).<\/span>
\nSolution: \\(\\Delta S = nC_p\\ln\\frac{T_2}{T_1} = \\frac{7}{2}R\\ln\\frac{340}{300}\\).<\/span><\/p>\n

12. Find the heat transfer in an isobaric expansion, \\(P = 2\\ \\text{atm}\\), \\(\\Delta V = 3\\ \\text{L}\\), \\(C_p = \\frac{7}{2}R\\).<\/span>
\nSolution: \\(Q = P\\Delta V + nC_p\\Delta T = 2 \\times 3 + \\frac{7}{2}R\\Delta T\\).<\/span><\/p>\n

13. Determine the work done in an isobaric process for \\(P = 4\\ \\text{bar}\\), \\(\\Delta V = 5\\ \\text{m}^3\\).<\/span>
\nSolution: \\(W = P\\Delta V = 4 \\times 5 = 20\\ \\text{bar\u00b7m}^3\\).<\/span><\/p>\n

14. Calculate the internal energy change for an isobaric compression, \\(n = 2\\ \\text{mol}\\), \\(C_v = \\frac{3}{2}R\\), \\(\\Delta T = -30\\ \\text{K}\\).<\/span>
\nSolution: \\(\\Delta U = nC_v\\Delta T = 2 \\cdot \\frac{3}{2}R \\cdot (-30) \\approx -753\\ \\text{J}\\).<\/span><\/p>\n

15. Find the entropy change in an isobaric process, \\(n = 1.5\\ \\text{mol}\\), \\(\\Delta T = 60\\ \\text{K}\\), \\(T_1 = 400\\ \\text{K}\\), \\(C_p = \\frac{5}{2}R\\).<\/span>
\nSolution: \\(\\Delta S = nC_p\\ln\\frac{T_2}{T_1} = 1.5 \\cdot \\frac{5}{2}R\\ln\\frac{460}{400}\\).<\/span><\/p>\n

16. Determine the heat transfer for an isobaric expansion, \\(P = 3\\ \\text{bar}\\), \\(\\Delta V = 2\\ \\text{m}^3\\), \\(C_p = \\frac{5}{2}R\\), \\(n = 2\\ \\text{mol}\\).<\/span>
\nSolution: \\(Q = P\\Delta V + nC_p\\Delta T = 3 \\times 2 + 2 \\cdot \\frac{5}{2}R\\Delta T\\).<\/span><\/p>\n

17. Calculate the work done on 3 moles of a gas undergoing an isobaric compression, \\(P = 5\\ \\text{atm}\\), \\(\\Delta V = -4\\ \\text{L}\\).<\/span>
\nSolution: \\(W = P\\Delta V = 5 \\times (-4) = -20\\ \\text{L\u00b7atm}\\).<\/span><\/p>\n

18. Determine the internal energy change for \\(n = 4\\ \\text{mol}\\), \\(C_v = \\frac{7}{2}R\\), \\(\\Delta T = 15\\ \\text{K}\\) in an isobaric process.<\/span>
\nSolution: \\(\\Delta U = nC_v\\Delta T = 4 \\cdot \\frac{7}{2}R \\cdot 15 \\approx 3157\\ \\text{J}\\).<\/span><\/p>\n

19. Find the heat transfer in an isobaric process, \\(P = 4\\ \\text{atm}\\), \\(\\Delta V = 5\\ \\text{L}\\), \\(n = 2\\ \\text{mol}\\), \\(C_p = \\frac{5}{2}R\\).<\/span>
\nSolution: \\(Q = P\\Delta V + nC_p\\Delta T = 4 \\times 5 + 2 \\cdot \\frac{5}{2}R\\Delta T\\).<\/span><\/p>\n

20. Determine the work done in an isobaric compression, \\(P = 7\\ \\text{bar}\\), \\(\\Delta V = -2\\ \\text{m}^3\\).<\/span>
\nSolution: \\(W = P\\Delta V = 7 \\times (-2) = -14\\ \\text{bar\u00b7m}^3\\).<\/span><\/p>\n

21. Calculate the internal energy change for 3 moles of an ideal gas undergoing an isobaric process, \\(C_v = \\frac{5}{2}R\\), \\(\\Delta T = 20\\ \\text{K}\\).<\/span>
\nSolution: \\(\\Delta U = nC_v\\Delta T = 3 \\cdot \\frac{5}{2}R \\cdot 20 \\approx 1256\\ \\text{J}\\).<\/span><\/p>\n

22. Find the entropy change for an isobaric expansion, \\(n = 1\\ \\text{mol}\\), \\(C_p = \\frac{7}{2}R\\), \\(\\Delta T = 30\\ \\text{K}\\), \\(T_1 = 250\\ \\text{K}\\).<\/span>
\nSolution: \\(\\Delta S = nC_p\\ln\\frac{T_2}{T_1} = \\frac{7}{2}R\\ln\\frac{280}{250}\\).<\/span><\/p>\n

23. Determine the heat transfer in an isobaric process, \\(P = 6\\ \\text{bar}\\), \\(\\Delta V = 4\\ \\text{m}^3\\), \\(n = 3\\ \\text{mol}\\), \\(C_p = \\frac{3}{2}R\\).<\/span>
\nSolution: \\(Q = P\\Delta V + nC_p\\Delta T = 6 \\times 4 + 3 \\cdot \\frac{3}{2}R\\Delta T\\).<\/span><\/p>\n

24. Calculate the work done by the system in an isobaric expansion with \\(P = 8\\ \\text{bar}\\), \\(\\Delta V = 3\\ \\text{m}^3\\).<\/span>
\nSolution: \\(W = P\\Delta V = 8 \\times 3 = 24\\ \\text{bar\u00b7m}^3\\).<\/span><\/p>\n

25. Determine the internal energy change for an isobaric process where \\(n = 2\\ \\text{mol}\\), \\(C_v = \\frac{7}{2}R\\), \\(\\Delta T = -10\\ \\text{K}\\).<\/span>
\nSolution: \\(\\Delta U = nC_v\\Delta T = 2 \\cdot \\frac{7}{2}R \\cdot (-10) \\approx -878\\ \\text{J}\\).<\/span><\/p>\n

26. Find the entropy change for a diatomic ideal gas in an isobaric compression, \\(n = 1.5\\ \\text{mol}\\), \\(T_1 = 350\\ \\text{K}\\), \\(\\Delta T = -40\\ \\text{K}\\).<\/span>
\nSolution: \\(\\Delta S = nC_p\\ln\\frac{T_2}{T_1} = 1.5 \\cdot \\frac{7}{2}R\\ln\\frac{310}{350}\\).<\/span><\/p>\n

27. Determine the heat transfer for 2 moles of a gas undergoing an isobaric expansion, \\(P = 5\\ \\text{bar}\\), \\(\\Delta V = 6\\ \\text{m}^3\\), \\(C_p = \\frac{5}{2}R\\).<\/span>
\nSolution: \\(Q = P\\Delta V + nC_p\\Delta T = 5 \\times 6 + 2 \\cdot \\frac{5}{2}R\\Delta T\\).<\/span><\/p>\n

28. Calculate the work done on the system in an isobaric compression with \\(P = 9\\ \\text{atm}\\), \\(\\Delta V = -3\\ \\text{L}\\).<\/span>
\nSolution: \\(W = P\\Delta V = 9 \\times (-3) = -27\\ \\text{L\u00b7atm}\\).<\/span><\/p>\n

29. Determine the internal energy change for 3 moles of a gas undergoing an isobaric process, \\(C_v = \\frac{3}{2}R\\), \\(\\Delta T = 15\\ \\text{K}\\).<\/span>
\nSolution: \\(\\Delta U = nC_v\\Delta T = 3 \\cdot \\frac{3}{2}R \\cdot 15 \\approx 564\\ \\text{J}\\).<\/span><\/p>\n

30. Find the entropy change in an isobaric expansion, \\(n = 4\\ \\text{mol}\\), \\(C_p = \\frac{5}{2}R\\), \\(\\Delta T = 25\\ \\text{K}\\), \\(T_1 = 300\\ \\text{K}\\).<\/span>
\nSolution: \\(\\Delta S = nC_p\\ln\\frac{T_2}{T_1} = 4 \\cdot \\frac{5}{2}R\\ln\\frac{325}{300}\\).<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

30 Isobaric thermodynamics processes – problems and solutions 1. PV diagram below shows an ideal gas undergoes an isobaric process. Calculate the work is done by the gas in the process AB. Known : Pressure (P) = 5 x 105 N\/m2 Initial volume (V1) = 2 m3 Final volume (V2) = 6 m3 Wanted : … 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":"Isobaric thermodynamics 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-1569","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\/1569","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=1569"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1569\/revisions"}],"predecessor-version":[{"id":9022,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1569\/revisions\/9022"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1569"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1569"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1569"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}