A gas contained in a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where p1 =1bar, V1 = 1m3, U1 =400kJ and p * 2 = 10 bar, V2 = 0.1 m3, U2 = 450 kJ: Process A: Constant-volume process from state 1 to a pressure of 10 bar, followed by a constantpressure process to state 2. Process B: Process from 1 to 2 during which the pressure-volume relation is pV = constant. Kinetic and potential effects can be ignored. For each of the processes A and B, (a) sketch the process on p-V coordinates, (b) evaluate the work, in kJ, and (c) evaluate the heat transfer, in kJ.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Alright, let\u2019s work through this carefully.<\/p>\n\n\n\n So, Work for Process A<\/strong> = (-900) kJ<\/strong><\/p>\n<\/blockquote>\n\n\n\n Since ( pV = \\text{constant} ), it is a polytropic process<\/strong> with ( n = 1 ).<\/p>\n\n\n\n Work formula for ( n = 1 ) process: Substituting: So, Work for Process B<\/strong> = (-230.26) kJ<\/strong><\/p>\n<\/blockquote>\n\n\n\n From First Law of Thermodynamics<\/strong>: [ [ This problem involves analyzing two different thermodynamic processes that take a gas between the same starting and ending conditions, but using different paths. In Process A, the gas first undergoes a constant volume compression, meaning the piston is locked, and the pressure increases from 1 bar to 10 bar without any volume change. Since the volume doesn\u2019t change, no work is done during this step (( W = 0 )). After reaching 10 bar, the gas is compressed at constant pressure from a volume of 1 m\u00b3 to 0.1 m\u00b3. For a constant pressure process, the work is simply the pressure multiplied by the volume change. Here, because the volume is decreasing, the work comes out negative, meaning the system is compressed.<\/p>\n\n\n\n In Process B, the gas undergoes a process where ( pV ) remains constant. This is known as a polytropic process with ( n = 1 ). The work done during this process involves integrating ( p \\, dV ) while keeping ( pV = \\text{constant} ), leading to the logarithmic formula for work. The work is again negative because the volume is decreasing during compression, although it is a smaller magnitude compared to Process A.<\/p>\n\n\n\n Once the work is known for each process, we use the first law of thermodynamics to find the heat transfer: ( Q = \\Delta U + W ). The internal energy change ( \\Delta U ) is positive, meaning the gas absorbs internal energy, but because the work terms are negative and larger in magnitude, the heat transfer ends up being negative for both processes. A negative heat transfer indicates that the system loses heat to the surroundings.<\/p>\n\n\n\n The results show that Process A requires more compression work (\u2013900 kJ) compared to Process B (\u2013230.26 kJ), and correspondingly, loses more heat. This highlights how the path between states significantly affects work and heat interactions even if the initial and final states are identical.<\/p>\n\n\n\n A gas contained in a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where p1 =1bar, V1 = 1m3, U1 =400kJ and p * 2 = 10 bar, V2 = 0.1 m3, U2 = 450 kJ: Process A: Constant-volume process from state 1 to a pressure of […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-209207","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209207","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=209207"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209207\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209207"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209207"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209207"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\nFirst, given data<\/strong>:<\/h3>\n\n\n\n
\n
( p_1 = 1\\, \\text{bar} = 100\\, \\text{kPa} )
( V_1 = 1\\, \\text{m}^3 )
( U_1 = 400\\, \\text{kJ} )<\/li>\n\n\n\n
( p_2 = 10\\, \\text{bar} = 1000\\, \\text{kPa} )
( V_2 = 0.1\\, \\text{m}^3 )
( U_2 = 450\\, \\text{kJ} )<\/li>\n<\/ul>\n\n\n\n
\n\n\n\n(a) Sketch on p-V Diagram<\/strong>:<\/h2>\n\n\n\n
\n
\n\n\n\n(b) Work done (W)<\/strong><\/h2>\n\n\n\n
Process A:<\/h3>\n\n\n\n
\n
( W = 0 ) (because ( dV = 0 ))<\/li>\n\n\n\n
[
W = p (V_2 – V_1)
]
Substituting values:
[
W = 1000\\, \\text{kPa} \\times (0.1 – 1) \\, \\text{m}^3 = 1000 \\times (-0.9) = -900\\, \\text{kJ}
]<\/li>\n<\/ul>\n\n\n\n\n
\n\n\n\nProcess B:<\/h3>\n\n\n\n
[
W = p_1 V_1 \\ln\\left( \\frac{V_2}{V_1} \\right)
]<\/p>\n\n\n\n
[
W = (100\\, \\text{kPa})(1\\, \\text{m}^3) \\ln\\left( \\frac{0.1}{1} \\right)
]
[
= 100 \\times \\ln(0.1)
]
[
= 100 \\times (-2.3026)
]
[
= -230.26\\, \\text{kJ}
]<\/p>\n\n\n\n\n
\n\n\n\n(c) Heat transfer (Q)<\/strong><\/h2>\n\n\n\n
[
Q = \\Delta U + W
]
where:
[
\\Delta U = U_2 – U_1 = 450 – 400 = 50\\, \\text{kJ}
]<\/p>\n\n\n\n
\n\n\n\nProcess A:<\/h3>\n\n\n\n
Q_A = \\Delta U + W_A
]
[
Q_A = 50 + (-900)
]
[
Q_A = -850\\, \\text{kJ}
]<\/p>\n\n\n\n
\n\n\n\nProcess B:<\/h3>\n\n\n\n
Q_B = \\Delta U + W_B
]
[
Q_B = 50 + (-230.26)
]
[
Q_B = -180.26\\, \\text{kJ}
]<\/p>\n\n\n\n
\n\n\n\nSummary of Results:<\/strong><\/h1>\n\n\n\n
<\/th> Work (kJ)<\/th> Heat Transfer (kJ)<\/th><\/tr><\/thead> Process A<\/td> -900<\/td> -850<\/td><\/tr> Process B<\/td> -230.26<\/td> -180.26<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
\n\n\n\nExplanation (300+ words):<\/strong><\/h2>\n\n\n\n
\n","protected":false},"excerpt":{"rendered":"