{"id":5223,"date":"2021-11-18T17:39:21","date_gmt":"2021-11-18T12:09:21","guid":{"rendered":"https:\/\/acejee.com\/blog\/?p=5223"},"modified":"2024-09-30T08:10:39","modified_gmt":"2024-09-30T02:40:39","slug":"fluids-surface-energy-iit-jee-jee-mains","status":"publish","type":"post","link":"https:\/\/acejee.com\/blog\/fluids-surface-energy-iit-jee-jee-mains\/","title":{"rendered":"Fluids | Surface Energy | IIT JEE & JEE Mains"},"content":{"rendered":"\t\t
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A molecule on the surface has greater potential energy than a molecule inside the liquid. The extra energy in the surface layer is called surface energy. Here we will examine the surface energy of a bubble<\/a> and of a liquid drop<\/a> of surface tension $T$ and radius $r$, followed by a review of ‘coalescing of air bubbles<\/a> and radii of resulting bubble’ under isothermal conditions. Then we will determine the radius of curvature of the common surface<\/a> between two bubbles, followed by the determination of change in surface energy as liquid drops coalesce<\/a> together. In the end, we will look at the energy required to enlarge a bubble<\/a>.<\/p>

So, starting with air bubble<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Surface Energy of Air Bubble<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Air\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Let’s say that the pressure inside the bubble is $p_i$ and outside it, is $p_o$. So difference in pressure $\\Delta p$ $=p_i – p_o$<\/p>

Now, if you consider half the bubble, as shown, net force in say $x$ direction (i.e. in the horizontal direction), $F_{net, x}$ must be zero, i.e.
$F_{net,x}= \\Delta p \\pi R^2$ $-2 \\times T \\times 2 \\pi R$ $=0$<\/p>

or, $\\Delta p = \\cfrac{4T}{R}$
So, how much work will be done in increasing the radii from say $r = 0$ to $r = R$<\/p>

$W = \\int dW$ $= \\int p dV$ $= \\int_0^R \\cfrac{4T}{r} 4 \\pi r^2 dr$ $= 2T.4 \\pi R^2$ $= T (2A)$, where $A = 4 \\pi R^2$<\/p>

This work done is stored as the surface energy $U$
So, $U = T.2A$
and surface energy per unit area $=\\cfrac{U}{2A} = T$<\/p>

Note, we have $2A$ because there are two surfaces, each of surface area ($\\approx 4 \\pi R^2$) in case of a bubble.\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Surface Energy of Liquid Drop<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Liquid\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Let’s say that the pressure inside the drop is $p_i$ and outside it, is $p_o$. So difference in pressure $\\Delta p$ $=p_i – p_o$<\/p>

Now, if you consider half the drop,
$\\Delta p \\pi R^2$ $=T \\times 2 \\pi R$
or, $\\Delta p = \\cfrac{2T}{R}$
So, how much work will be done in increasing the radii from say $r = 0$ to $r = R$<\/p>

$W = \\int dW$ $= \\int p dV$ $= \\int_0^R \\cfrac{2T}{r} 4 \\pi r^2 dr$ $= T.4 \\pi R^2$ $= T (A)$, where $A = 4 \\pi R^2$<\/p>

This work done is stored as the surface energy $U$,
So, $U = T.A$
and surface energy per unit surface area $=T$<\/p>

Note, unlike bubble, drop has only 1 surface with area $=4 \\pi R^2$.\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Coalescing of Bubbles | Radii of resulting bubble<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Coalescing\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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For two bubbles in vacuum, with radii $r_1$ and $r_2$, number of moles $n_1$ and $n_2$, pressure $p_1$ and $p_2$, and volume $V_1$ and $V_2$ assuming that they coalesce under isothermal conditions to form a bubble of radius $r$, number of moles $n$ ($=n_1 + n_2$), pressure $p$, and volume $V$.<\/p>

Now, assuming that the gas \/ air in the bubbles behaves like an ideal gas, meaning the gas follows the ideal gas law: $pV = nRT$ or $\\cfrac{pV}{RT} = n$<\/p>

Then, it follows that $\\cfrac{p_1V_1}{RT} + \\cfrac{p_2V_2}{RT} = \\cfrac{pV}{RT}$ [$n_1 + n_2 = n$ ]<\/p>

Or, $p_1 r_1^3 + p_2 r_2^3 = p r^2$<\/p>

Now, from earlier, for a bubble of radius $r$, $\\Delta p = p_i – p_o = \\cfrac{4T}{r}$,<\/p>

In vacuum this reduces to $p_i = \\cfrac{4T}{r}$<\/p>

or $4T r_1^2 + 4T r_2^2 = 4T r^2$<\/p>

or $r_1^2 + r_2^2 = r^2$\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Radius of curvature of the common surface between two bubbles<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Radius\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Let’s say that the pressure on the two sides of the common surface is $p_1$ and $p_2$<\/p>

Then the forces on it along its axis (i.e. in the vertical direction) are:
$2T \\times 2 \\pi r \\sin \\theta$ : in the downward direction due to surface tension $T$
$|p_1 – p_2| \\pi r^2$ : in the upward direction due to pressure difference<\/p>

And since the net force on this surface is zero (as it is in equilibrium),
$2T \\times 2 \\pi r \\sin \\theta$ $=\\Delta p \\pi r^2$<\/p>

or, $\\Delta p = \\cfrac{4T}{(r\/ \\sin \\theta)}$ $= \\cfrac{4T}{R}$<\/p>

or, $R = \\cfrac{4T}{\\Delta p}$\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Coalescing of liquid drops and resulting change in surface energy<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Coalescing\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Let’s say that the two drops (of incompressible liquid) had a radius of $r_1$ and $r_2$<\/p>

So, the total surface energy before coalescing is $=T.4 \\pi r_1^2 + T.4 \\pi r_2^2$ $=4 \\pi T (r_1^2 + r_2^2)$<\/p>

After coalescing together, the radius of the resulting drop is given by $V = V_1 + V_2$
i.e. $r = (r_1^3 + r_2^3)^{1\/3}$<\/p>

So the surface energy of the resulting drop $=4 \\pi T (r_1^3 + r_2^3)^{2\/3}$<\/p>

Hence, drop in surface energy $=4 \\pi T (r_1^2 + r_2^2 – (r_1^3 + r_2^3)^{2\/3})$<\/p>

This difference in energy would show up as the heat energy which will raise the temperature of the drop.<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t

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Energy required to enlarge a bubble<\/h2>\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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\n\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t\"Energy\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t
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Let’s say that the initial radius of the bubble was $r_i$ and final radius is $r_f$<\/p>

So, work needed $=$ increase in surface energy $=$ $2T.4 \\pi (r_f^2 – r_i^2)$ $=8 \\pi T (r_f^2 – r_i^2)$\u00a0<\/p>\t\t\t\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/div>\n\t\t\t\t\t<\/div>\n\t\t<\/section>\n\t\t\t\t<\/div>\n\t\t","protected":false},"excerpt":{"rendered":"

A molecule in the surface has greater potential energy than a molecule inside the liquid. The extra energy in the surface layer is called the surface energy<\/p>\n","protected":false},"author":1,"featured_media":5225,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"ocean_post_layout":"","ocean_both_sidebars_style":"","ocean_both_sidebars_content_width":0,"ocean_both_sidebars_sidebars_width":0,"ocean_sidebar":"0","ocean_second_sidebar":"0","ocean_disable_margins":"enable","ocean_add_body_class":"","ocean_shortcode_before_top_bar":"","ocean_shortcode_after_top_bar":"","ocean_shortcode_before_header":"","ocean_shortcode_after_header":"","ocean_has_shortcode":"","ocean_shortcode_after_title":"","ocean_shortcode_before_footer_widgets":"","ocean_shortcode_after_footer_widgets":"","ocean_shortcode_before_footer_bottom":"","ocean_shortcode_after_footer_bottom":"","ocean_display_top_bar":"default","ocean_display_header":"default","ocean_header_style":"","ocean_center_header_left_menu":"0","ocean_custom_header_template":"0","ocean_custom_logo":0,"ocean_custom_retina_logo":0,"ocean_custom_logo_max_width":0,"ocean_custom_logo_tablet_max_width":0,"ocean_custom_logo_mobile_max_width":0,"ocean_custom_logo_max_height":0,"ocean_custom_logo_tablet_max_height":0,"ocean_custom_logo_mobile_max_height":0,"ocean_header_custom_menu":"0","ocean_menu_typo_font_family":"0","ocean_menu_typo_font_subset":"","ocean_menu_typo_font_size":0,"ocean_menu_typo_font_size_tablet":0,"ocean_menu_typo_font_size_mobile":0,"ocean_menu_typo_font_size_unit":"px","ocean_menu_typo_font_weight":"","ocean_menu_typo_font_weight_tablet":"","ocean_menu_typo_font_weight_mobile":"","ocean_menu_typo_transform":"","ocean_menu_typo_transform_tablet":"","ocean_menu_typo_transform_mobile":"","ocean_menu_typo_line_height":0,"ocean_menu_typo_line_height_tablet":0,"ocean_menu_typo_line_height_mobile":0,"ocean_menu_typo_line_height_unit":"","ocean_menu_typo_spacing":0,"ocean_menu_typo_spacing_tablet":0,"ocean_menu_typo_spacing_mobile":0,"ocean_menu_typo_spacing_unit":"","ocean_menu_link_color":"","ocean_menu_link_color_hover":"","ocean_menu_link_color_active":"","ocean_menu_link_background":"","ocean_menu_link_hover_background":"","ocean_menu_link_active_background":"","ocean_menu_social_links_bg":"","ocean_menu_social_hover_links_bg":"","ocean_menu_social_links_color":"","ocean_menu_social_hover_links_color":"","ocean_disable_title":"on","ocean_disable_heading":"default","ocean_post_title":"","ocean_post_subheading":"","ocean_post_title_style":"","ocean_post_title_background_color":"","ocean_post_title_background":0,"ocean_post_title_bg_image_position":"","ocean_post_title_bg_image_attachment":"","ocean_post_title_bg_image_repeat":"","ocean_post_title_bg_image_size":"","ocean_post_title_height":0,"ocean_post_title_bg_overlay":0.5,"ocean_post_title_bg_overlay_color":"","ocean_disable_breadcrumbs":"off","ocean_breadcrumbs_color":"","ocean_breadcrumbs_separator_color":"","ocean_breadcrumbs_links_color":"","ocean_breadcrumbs_links_hover_color":"","ocean_display_footer_widgets":"default","ocean_display_footer_bottom":"default","ocean_custom_footer_template":"0","cybocfi_hide_featured_image":"yes","ocean_post_oembed":"","ocean_post_self_hosted_media":"","ocean_post_video_embed":"","ocean_link_format":"","ocean_link_format_target":"self","ocean_quote_format":"","ocean_quote_format_link":"post","ocean_gallery_link_images":"off","ocean_gallery_id":[],"footnotes":""},"categories":[16],"tags":[],"class_list":["post-5223","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-iit-jee-physics-chapter-wise-approach","entry","has-media"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/5223","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/comments?post=5223"}],"version-history":[{"count":1,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/5223\/revisions"}],"predecessor-version":[{"id":9434,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/posts\/5223\/revisions\/9434"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/media\/5225"}],"wp:attachment":[{"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/media?parent=5223"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/categories?post=5223"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/acejee.com\/blog\/wp-json\/wp\/v2\/tags?post=5223"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}