{"id":1565,"date":"2018-03-10T13:48:33","date_gmt":"2018-03-10T05:48:33","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1565"},"modified":"2023-08-19T01:46:20","modified_gmt":"2023-08-19T01:46:20","slug":"isochoric-thermodynamics-processes-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/isochoric-thermodynamics-processes-problems-and-solutions.htm","title":{"rendered":"Isochoric thermodynamics processes – problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

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

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

\"IsochoricSolution :<\/span><\/p>\n

Process AB is an isochoric process <\/a>(constant volume). The volume is constant so that no work is done by the gas.<\/span><\/p>\n

.<\/span><\/p>\n

2. Three moles of monoatomic gas at 47o<\/sup>C and at pressure<\/a> 2 x 105<\/sup> Pa, undergoes isochoric process so that pressure increases 3 x 105<\/sup> Pa. The change in internal energy of the gas is… Universal gas constant (R) = 8.315 J\/mol.K<\/span><\/p>\n

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

Initial temperature <\/a>(T1<\/sub>) = 47o<\/sup>C + 273 = 320 K<\/span><\/p>\n

Initial pressure (P1<\/sub>) = 2 x 105<\/sup> Pa<\/span><\/p>\n

Final pressure (P2<\/sub>) = 3 x 105<\/sup> Pa<\/span><\/p>\n

Universal gas constant (R) = 8.315 J\/mol.K<\/span><\/p>\n

Number of moles (n) = 3<\/span><\/p>\n

Wanted:<\/u> The change in internal energy of the gas.<\/span><\/p>\n

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

In the isochoric process, the volume is kept constant so that no work is done by the gas (W = 0).<\/span><\/p>\n

The first law of thermodynamics<\/a> :<\/b><\/span><\/p>\n

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

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

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

\u0394U = internal energy, Q = heat<\/i><\/span><\/p>\n

Internal energy of gas :<\/u><\/span><\/p>\n

\u0394U = 3\/2 n R \u0394T = 3\/2 n R (T2<\/sub> \u2013 T1<\/sub>)<\/span><\/p>\n

Gay-Lussac<\/u>‘s law<\/u> (<\/u>constant volume<\/u>)<\/u><\/a> : <\/span><\/p>\n

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

The change in internal energy of gas :<\/span><\/p>\n

\u0394U = 3\/2 n R (T2<\/sub> \u2013 T1<\/sub>) = 3\/2 (3)(8.315)(480-320)<\/span><\/p>\n

\u0394U = 3\/2 (24.945)(160) = 3\/2 (3991.2)<\/span><\/p>\n

\u0394U = 5986.8 Joule<\/span><\/p>\n

3. 0.2 moles of monatomic gases at 27o<\/sup>C are in a closed container. The heat<\/a> is added to the gas so that temperature of gas becomes 400 K is… Universal gas constant (R) = 8.315 J\/mol.K<\/span><\/p>\n

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

Number of moles (n) = 0.2 mol<\/span><\/p>\n

Initial temperature (T1<\/sub>) = 27o<\/sup>C + 273 = 300 K <\/span><\/p>\n

Final temperature (T2<\/sub>) = 400 K<\/span><\/p>\n

Universal constant gas (R) = 8.315 J\/mol.K <\/span><\/p>\n

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

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

In isochoric process, volume is kept constant so that no work is done by the gas (W = 0).<\/span><\/p>\n

The first law of thermodynamics :<\/b><\/span><\/p>\n

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

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

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

\u0394U = internal energy, Q = heat<\/i><\/span><\/p>\n

The internal energy of gas :<\/span><\/p>\n

\u0394U = 3\/2 n R \u0394T = 3\/2 n R (T2<\/sub> \u2013 T1<\/sub>)<\/span><\/p>\n

\u0394U = 3\/2 (0.2)(8.315)(400-300) <\/span><\/p>\n

\u0394U = 3\/2 (0.2)(8.315)(100)<\/span><\/p>\n

\u0394U = 249.45 Joule<\/span><\/p>\n

4. Calculate the heat transfer for an ideal gas undergoing an isochoric process from an initial temperature of 300 K to a final temperature of 400 K. Assume 2 mol of gas, and the molar heat capacity at constant volume (C\u1d65) is 20 J\/(mol\u00b7K).<\/span>
\nSolution: \u0394Q = n \u00d7 C\u1d65 \u00d7 \u0394T = 2 mol \u00d7 20 J\/(mol\u00b7K) \u00d7 (400 K – 300 K) = 4000 J<\/span><\/p>\n

5. Find the change in internal energy for the above problem.<\/span>
\nSolution: \u0394U = \u0394Q = 4000 J<\/span><\/p>\n

6. Determine the work done on a system during an isochoric process for the above conditions.<\/span>
\nSolution: W = 0 J (since the volume doesn’t change, no work is done)<\/span><\/p>\n

7. For a monatomic ideal gas undergoing an isochoric process, if the initial pressure is 2 atm and the final pressure is 3 atm, what is the ratio of final to initial temperatures?<\/span>
\nSolution: Since P\u2081\/T\u2081 = P\u2082\/T\u2082, T\u2082\/T\u2081 = 3\/2<\/span><\/p>\n

8. What is the entropy change for an ideal gas in an isochoric process when the temperature changes from 300 K to 600 K, and n = 2 mol, C\u1d65 = 20 J\/(mol\u00b7K)?<\/span>
\nSolution: \u0394S = n \u00d7 C\u1d65 \u00d7 ln(T\u2082\/T\u2081) = 2 \u00d7 20 \u00d7 ln(600\/300) \u2248 27.73 J\/K<\/span><\/p>\n

9. If the initial state of a diatomic ideal gas is defined by V = 2 L, P = 1 atm, and T = 300 K, find the final pressure if the temperature is doubled in an isochoric process.<\/span>
\nSolution: P\u2082 = 2 \u00d7 P\u2081 = 2 atm<\/span><\/p>\n

10. Find the change in Gibbs free energy for an isochoric process.<\/span>
\nSolution: \u0394G = 0 (For an isochoric process in a closed system, \u0394G = 0)<\/span><\/p>\n

11. Calculate the final temperature of an ideal gas undergoing an isochoric process if the initial temperature is 200 K, and the initial and final pressures are 2 atm and 4 atm, respectively.<\/span>
\nSolution: T\u2082 = 2 \u00d7 T\u2081 = 400 K<\/span><\/p>\n

12. For an ideal gas, if the heat capacity at constant volume (C\u1d65) is 30 J\/(mol\u00b7K), find the heat transfer when the temperature changes from 300 K to 450 K, with 3 mol of gas.<\/span>
\nSolution: \u0394Q = n \u00d7 C\u1d65 \u00d7 \u0394T = 3 \u00d7 30 \u00d7 150 = 13500 J<\/span><\/p>\n

13. For the same process as above, calculate the change in internal energy.<\/span>
\nSolution: \u0394U = \u0394Q = 13500 J<\/span><\/p>\n

14. Determine the entropy change for an isochoric process with n = 1 mol, C\u1d65 = 25 J\/(mol\u00b7K), T\u2081 = 200 K, and T\u2082 = 400 K.<\/span>
\nSolution: \u0394S = n \u00d7 C\u1d65 \u00d7 ln(T\u2082\/T\u2081) = 25 \u00d7 ln(2) \u2248 17.33 J\/K<\/span><\/p>\n

15. Find the work done by the system during an isochoric process of 3 mol of gas, and the temperature changes from 200 K to 300 K.<\/span>
\nSolution: W = 0 J (since the volume doesn’t change, no work is done)<\/span><\/p>\n

16. Calculate the heat transfer for an ideal gas undergoing an isochoric process with an initial temperature of 150 K, final temperature of 300 K, and C\u1d65 = 15 J\/(mol\u00b7K) for 4 mol of gas.<\/span>
\nSolution: \u0394Q = n \u00d7 C\u1d65 \u00d7 \u0394T = 4 \u00d7 15 \u00d7 150 = 9000 J<\/span><\/p>\n

17. What is the entropy change for an ideal gas in an isochoric process with n = 1 mol, C\u1d65 = 30 J\/(mol\u00b7K), T\u2081 = 100 K, and T\u2082 = 200 K?<\/span>
\nSolution: \u0394S = n \u00d7 C\u1d65 \u00d7 ln(T\u2082\/T\u2081) = 30 \u00d7 ln(2) \u2248 20.79 J\/K<\/span><\/p>\n

18. Determine the final pressure of a gas undergoing an isochoric process, given that P\u2081 = 5 atm, T\u2081 = 250 K, and T\u2082 = 500 K.<\/span>
\nSolution: P\u2082 = (T\u2082\/T\u2081) \u00d7 P\u2081 = 2 \u00d7 5 atm = 10 atm<\/span><\/p>\n

19. Find the heat transfer for 5 mol of a monatomic ideal gas undergoing an isochoric process from 300 K to 600 K. Assume C\u1d65 = 15 J\/(mol\u00b7K).<\/span>
\nSolution: \u0394Q = n \u00d7 C\u1d65 \u00d7 \u0394T = 5 \u00d7 15 \u00d7 300 = 22500 J<\/span><\/p>\n

20. What is the change in internal energy for the above problem?<\/span>
\nSolution: \u0394U = \u0394Q = 22500 J<\/span><\/p>\n

21. Determine the entropy change for an isochoric process where n = 2 mol, C\u1d65 = 25 J\/(mol\u00b7K), T\u2081 = 300 K, and T\u2082 = 600 K.<\/span>
\nSolution: \u0394S = n \u00d7 C\u1d65 \u00d7 ln(T\u2082\/T\u2081) = 2 \u00d7 25 \u00d7 ln(2) \u2248 34.66 J\/K<\/span><\/p>\n

22. Calculate the final temperature of 1 mol of a monatomic ideal gas undergoing an isochoric process if the initial temperature is 400 K, and the initial and final pressures are 3 atm and 6 atm, respectively.<\/span>
\nSolution: T\u2082 = 2 \u00d7 T\u2081 = 800 K<\/span><\/p>\n

23. For a diatomic ideal gas undergoing an isochoric process, calculate the change in internal energy when the temperature changes from 300 K to 600 K, with 2 mol of gas, and C\u1d65 = 30 J\/(mol\u00b7K).<\/span>
\nSolution: \u0394U = n \u00d7 C\u1d65 \u00d7 \u0394T = 2 \u00d7 30 \u00d7 300 = 18000 J<\/span><\/p>\n

24. Calculate the heat transfer for an ideal gas undergoing an isochoric process with an initial temperature of 100 K, final temperature of 300 K, and C\u1d65 = 20 J\/(mol\u00b7K) for 2 mol of gas.<\/span>
\nSolution: \u0394Q = n \u00d7 C\u1d65 \u00d7 \u0394T = 2 \u00d7 20 \u00d7 200 = 8000 J<\/span><\/p>\n

25. Find the work done on the system during an isochoric process for the above conditions.<\/span>
\nSolution: W = 0 J (since the volume doesn’t change, no work is done)<\/span><\/p>\n

26. What is the entropy change for an ideal gas in an isochoric process when the temperature changes from 400 K to 800 K, and n = 3 mol, C\u1d65 = 20 J\/(mol\u00b7K)?<\/span>
\nSolution: \u0394S = n \u00d7 C\u1d65 \u00d7 ln(T\u2082\/T\u2081) = 3 \u00d7 20 \u00d7 ln(2) \u2248 41.58 J\/K<\/span><\/p>\n

27. Find the change in Gibbs free energy for an isochoric process.<\/span>
\nSolution: \u0394G = 0 (For an isochoric process in a closed system, \u0394G = 0)<\/span><\/p>\n

28. Determine the final pressure of a gas undergoing an isochoric process, given that P\u2081 = 3 atm, T\u2081 = 300 K, and T\u2082 = 450 K.<\/span>
\nSolution: P\u2082 = (T\u2082\/T\u2081) \u00d7 P\u2081 = 1.5 \u00d7 3 atm = 4.5 atm<\/span><\/p>\n

29. Calculate the change in internal energy for a system undergoing an isochoric process with 3 mol of gas, C\u1d65 = 20 J\/(mol\u00b7K), and the temperature changes from 200 K to 400 K.<\/span>
\nSolution: \u0394U = n \u00d7 C\u1d65 \u00d7 \u0394T = 3 \u00d7 20 \u00d7 200 = 12000 J<\/span><\/p>\n

30. Determine the entropy change for an isochoric process where n = 4 mol, C\u1d65 = 30 J\/(mol\u00b7K), T\u2081 = 150 K, and T\u2082 = 300 K.<\/span>
\nSolution: \u0394S = n \u00d7 C\u1d65 \u00d7 ln(T\u2082\/T\u2081) = 4 \u00d7 30 \u00d7 ln(2) \u2248 55.86 J\/K<\/span><\/p>\n

These problems cover various concepts related to isochoric processes, such as heat transfer, internal energy change, work done, entropy change, and more.<\/span><\/p>\n","protected":false,"gt_translate_keys":[{"key":"rendered","format":"html"}]},"excerpt":{"rendered":"

30 Isochoric thermodynamics processes – problems and solutions 1. PV diagram below shows an ideal gas undergoes an isochoric process. Calculate the work is done by the gas in the process AB. Solution : Process AB is an isochoric process (constant volume). The volume is constant so that no work is done by the gas. … 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":"Isochoric 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-1565","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\/1565","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=1565"}],"version-history":[{"count":2,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1565\/revisions"}],"predecessor-version":[{"id":9023,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/posts\/1565\/revisions\/9023"}],"wp:attachment":[{"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/media?parent=1565"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/categories?post=1565"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gurumuda.net\/physics\/wp-json\/wp\/v2\/tags?post=1565"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}