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id="main-column"> <h1 class="page_title"> Processes <a href="#" class="mark-page-favorite pull-right" data-pid="852" title="Mark as favorite" onclick="return false;"><i class="fa fa-star-o"></i></a> </h1> <ol class="breadcrumb"> <li><a href="../../../physics.html"><i class="fa fa-home"></i></a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><a href="../849/engineering.html">Engineering</a><i class="fa fa-fw fa-chevron-right divider"></i></li><li><span class="gray">Processes</span></li> <span class="pull-right" style="color: #555" title="Suggested study time: 10 minutes"><i class="fa fa-clock-o"></i> 10'</span> </ol> <article id="main-article"> <p><img alt="" src="../../engineering/tyre-1.jpg" style="float: left; width: 250px; height: 166px;"></p> <p>To avoid needlessly long descriptions, we can use diagrams to show the processes that take place in a container of gas.</p> <hr class="hidden-separator"> <div class="panel panel-turquoise panel-has-colored-body"> <div class="panel-heading"> <div> <p>Key Concepts</p> </div> </div> <div class="panel-body"> <div> <p>A graph of pressure vs volume (<span class="math-tex">\(pV\)</span> diagram) enables us to show all possible types of process because of the interrelationship between pressure, volume and temperature in the ideal gas law: <span class="math-tex">\(pV\propto T\)</span></p> <p>We can use the first law to consider the changing heat, internal energy and work done: <span class="math-tex">\(Q=\Delta U+p\Delta V\)</span></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/164055813"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Essentials</p> </div> </div> <div class="panel-body"> <div class="panel panel-has-colored-body panel-has-border panel-yellow"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Isovolumetric</p> </div> </div> <div class="panel-body"> <div> <p><img alt="" src="../../engineering/isovolumetric.png" style="float: right; width: 250px; height: 230px;">An <i>isovolumetric </i>process takes place at constant volume. This is represented by a vertical line (isochore) on a <span class="math-tex">\(pV\)</span> diagram.</p> <p>In this example, <span class="math-tex">\(\textrm{B}\to\textrm{C}\)</span> represents an increase in pressure and temperature.</p> <div class="box"> <p style="text-align: center;"><span class="math-tex">\(p\propto T\)</span></p> <ul> <li><span class="math-tex">\(p\)</span> is pressure (Pa)</li> <li><span class="math-tex">\(T\)</span> is temperature (K)</li> </ul> <p style="text-align: center;"><span class="math-tex">\({p_1\over T_1}={p_2\over T_2}\)</span></p> </div> <p>Since internal energy is increasing, heat must be gained by the gas. No work is done.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/355174980"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Isobaric</p> </div> </div> <div class="panel-body"> <div> <p><img alt="" src="../../engineering/isobar.png" style="float: left; width: 250px; height: 230px;">An <i>isobaric </i>process takes place at constant pressure. This is represented by a horizontal line (isobar) on a <span class="math-tex">\(pV\)</span>diagram.</p> <p>In this example, <span class="math-tex">\(\textrm{A}\to\textrm{B}\)</span> represents an expansion and increase in temperature.</p> <div class="box"> <p style="text-align: center;"><span class="math-tex">\(V\propto T\)</span></p> <ul> <li><span class="math-tex">\(V\)</span> is volume (m<sup>3</sup>)</li> <li><span class="math-tex">\(T\)</span> is temperature (K)</li> </ul> <p style="text-align: center;"><span class="math-tex">\({V_1\over T_1}={V_2\over T_2}\)</span></p> </div> <p>Since internal energy is increasing and work is done on the surroundings, heat must be gained by the gas.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/355174914"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Isothermal</p> </div> </div> <div class="panel-body"> <div> <p>A<img alt="" src="../../engineering/isotherm.png" style="float: right; width: 250px; height: 241px;">n <i>isothermal </i>process takes place at constant temperature. This is represented by a line of inverse proportion (isotherm) on a <span class="math-tex">\(pV\)</span> diagram.</p> <p>In this example, <span class="math-tex">\(\textrm{E}\to\textrm{F}\)</span> represents an expansion with a reduction in pressure.</p> <div class="box"> <p style="text-align: center;"><span class="math-tex">\(p\propto {1\over V}\)</span></p> <ul> <li><span class="math-tex">\(p\)</span> is pressure (Pa)</li> <li><span class="math-tex">\(V\)</span> is volume (m<sup>3</sup>)</li> </ul> <p style="text-align: center;"><span class="math-tex">\(p_1V_1=p_2V_2\)</span></p> </div> <p>Since internal energy is constant and work is done on the surroundings, heat must be gained by the gas. Isothermal processes are conducted slowly, to avoid increasing the kinetic energy of the particles.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/355174943"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> <div class="panel panel-has-colored-body panel-has-border panel-yellow panel-expandable"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Adiabatic</p> </div> </div> <div class="panel-body"> <div> <p><img alt="" src="../../engineering/adiabatic.png" style="float: left; width: 250px; height: 243px;">No heat is gained or lost during an <em>adiabatic</em> process. This is represented by an adiabat on a <span class="math-tex">\(pV\)</span> diagram, which has a steeper gradient throughout than an isotherm.</p> <p>In this example, <span class="math-tex">\(\textrm{G}\to\textrm{H}\)</span> represents an expansion with a reduction in pressure and temperature.</p> <div class="box"> <p>The following equation can be derived for monatomic gases:</p> <p style="text-align: center;"><span class="math-tex">\(p\propto {1\over V^{5\over 3}}\)</span></p> <ul> <li><span class="math-tex">\(p\)</span> is pressure (Pa)</li> <li><span class="math-tex">\(V\)</span> is volume (m<sup>3</sup>)</li> </ul> <p style="text-align: center;"><span class="math-tex">\(p_1V_1^{5\over 3}=p_2V_2^{5\over 3}\)</span></p> </div> <p>Since no heat enters or leaves, the decrease in internal energy must be equal to the work done on the surroundings. Adiabatic processes are conducted rapidly, to prevent sufficient time for heat to enter or leave.</p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/355175001"></iframe></div> </div> </div> <div class="panel-footer"> <div> <p> </p> </div> </div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-purple"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Summary</p> </div> </div> <div class="panel-body"> <div> <p><em>Watch this video to revise all of the possible first law changes.</em></p> <div class="video-embed vimeo"><iframe allow="accelerometer; autoplay; encrypted-media; gyroscope; picture-in-picture" allowfullscreen="" mozallowfullscreen="" webkitallowfullscreen="" height="420" width="100%" src="https://player.vimeo.com/video/355175302"></iframe></div> </div> </div> <div class="panel-footer"> <div> </div> </div> </div> <div class="panel panel-has-colored-body panel-green"> <div class="panel-heading"><a class="expander" href="#"><span class="fa fa-plus"></span></a> <div> <p>Test Yourself</p> </div> </div> <div class="panel-body"> <div> <p><em>Use flashcards to practise your recall.</em></p> <div class="tib-flashcard"><a class="show-flashcards btn btn-success btn-xs-block btn-block " data-levels="1" data-mode="Normal" data-topics="848" data-subject-id="6" data-n-flashcards="14" style="text-align:center">Show flashcards</a></div><hr> <p><em>Use quizzes to practise application of theory.</em> </p> <br><a class="btn btn-primary btn-block text-center" data-toggle="modal" href="#1e2b928e"><i class="fa fa-play"></i> START QUIZ!</a><div class="modal fade modal-slide-quiz" id="1e2b928e"> <div class="modal-dialog" style="width: 95vw; max-width: 960px"> <div class="modal-content"> <div class="modal-header slide-quiz-title"> <h4 class="modal-title" style="width: 100%;"> Thermal processes <strong class="q-number pull-right"> <span class="counter">1</span>/<span class="total">1</span> </strong> </h4> </div> <div class="modal-body p-xs-3"> <div class="slide-quiz" data-stats="6-302-852" style="opacity: 0"> <div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph shows two isochoric changes.</p><p><img alt="" height="297" src="../../engineering/screenshot-2019-07-30-at-08.05.47.png" width="370"></p><p>The ratio <span class="math-tex">\(Q_{AB} \over Q_{CD}\)</span> is</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(1\over2\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(4\)</span></span></label></p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(1\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(2\)</span></span></label></p></div><div class="q-explanation"><p>Since volume is constant, <span class="math-tex">\(Q = ΔU = {3\over2}nRΔT\Rightarrow Q\propto \Delta T\)</span></p><p><span class="math-tex">\(ΔT\)</span> is the same for both as the pressure triples in both cases (and <span class="math-tex">\(\Delta T\propto \Delta p\)</span>).</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph shows two isochoric changes. The amount of gas is such that <span class="math-tex">\(nR =10\)</span> kPa cm<sup>3 </sup>K<sup>-1</sup>.</p><p><img alt="" height="297" src="../../engineering/screenshot-2019-07-30-at-08.05.47.png" width="370"></p><p>The change in temperature from A to B is:</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>2000 K</span></label> </p><p><label class="radio"> <input type="radio"> <span>300 K</span></label> </p><p><label class="radio"> <input type="radio"> <span>200 K</span></label> </p><p><label class="radio"> <input type="radio"> <span>1000 K</span></label> </p></div><div class="q-explanation"><p>Recall the ideal gas law: <span class="math-tex">\(PV=nRT\)</span></p><p>This needs to be used separately at each point:</p><ul><li><span class="math-tex">\(T_A = (100 \times 100)\div 10\)</span></li><li><span class="math-tex">\(T_B = (100 \times 300)\div 10\)</span></li></ul></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph shows two isobaric changes.</p><p><img alt="" height="290" src="../../engineering/screenshot-2019-07-30-at-08.40.56.png" width="373"></p><p>The ratio <span class="math-tex">\(W_{AB}\over W_{CD}\)</span> is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(3\)</span></span></label></p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(1\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(1\over4\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(1\over3\)</span></span></label></p></div><div class="q-explanation"><p>Work done is equal to the area under the line.</p><p><span class="math-tex">\(W_{AB} = 200\times10^3 \times 150\times10^{-6} = 30\text{ J}\)</span></p><p><span class="math-tex">\(W_{CD} = 100\times10^3 \times 300\times10^{-6} = 30\text{ J}\)</span></p><p>NOTE: Including the unit prefixes was not required here as the question required a ratio.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph shows two isobaric changes.</p><p><img alt="" height="290" src="../../engineering/screenshot-2019-07-30-at-08.40.56.png" width="373"></p><p>The work done by the gas expanding from C to D is:</p></div><div class="q-answer"><p><label class="radio"> <input class="c" type="radio"> <span>30 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>30000 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>3 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>300 J</span></label> </p></div><div class="q-explanation"><p>Work done is equal to the area under the line.</p><p><span class="math-tex">\(W_{CD} = 100\times10^3 \times 300\times10^{-6} = 30\text{ J}\)</span></p><p>NOTE: Take care with unit conversions involving volume.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph shows two isobaric changes. The amount of gas is such that <span class="math-tex">\(nR =0.01\)</span> Pa m<sup>3 </sup>K<sup>-1</sup>.</p><p><img alt="" height="290" src="../../engineering/screenshot-2019-07-30-at-08.40.56.png" width="373"></p><p>The temperature change in each case is 3000K. How much heat is added expanding from C to D?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>30 J</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>75 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>45 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>15 J</span></label> </p></div><div class="q-explanation"><p>Internal energy: <span class="math-tex">\(ΔU = {3\over2} nRΔT = 45 \text{ J}\)</span></p><p>Work done by the gas in expansion: <span class="math-tex">\(W_{CD} = 100\times10^3 \times 300\times10^{-6} = 30\text{ J}\)</span></p><p>Using the first law of thermodynamics: <span class="math-tex">\(Q = ΔU + W\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph represents two isothermal changes.</p><p><img alt="" height="282" src="../../engineering/screenshot-2019-07-30-at-09.34.29.png" width="288"></p><p>The ratio <span class="math-tex">\(W_{AB}\over W_{CD}\)</span> is approximately:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>1</span></label> </p><p><label class="radio"> <input type="radio"> <span>3</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>4</span></label> </p><p><label class="radio"> <input type="radio"> <span>6</span></label> </p></div><div class="q-explanation"><p>Work done is equal to the area under the graph. In this case we need to count squares.</p><p><span class="math-tex">\(A\rightarrow B\)</span> 6.5 units</p><p><span class="math-tex">\(C\rightarrow D \)</span> 1.5 units</p><p><span class="math-tex">\(6.5\div1.5\approx4\)</span></p><p>NOTE: If required to calculate exact values for work done, you would need to calculate the work represented per square.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph represents two isothermal changes.</p><p><img alt="" height="282" src="../../engineering/screenshot-2019-07-30-at-09.34.29.png" width="288"></p><p>The ratio <span class="math-tex">\(Q_{AB}\over Q_{CD}\)</span> is approximately:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(6\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(1\over4\)</span></span></label></p><p><label class="radio"> <input class="c" type="radio"> <span><span class="math-tex">\(4\)</span></span></label></p><p><label class="radio"> <input type="radio"> <span><span class="math-tex">\(1\over 3\)</span></span></label></p></div><div class="q-explanation"><p>Since temperature is constant (isothermal), there is no change in internal energy. Using the first law, <span class="math-tex">\(Q=W\)</span></p><p>Work done is equal to the area under the graph. In this case we need to count squares.</p><p><span class="math-tex">\(A\rightarrow B\)</span> 6.5 units</p><p><span class="math-tex">\(C\rightarrow D \)</span> 1.5 units</p><p><span class="math-tex">\(6.5\div1.5\approx4\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph represents two isothermal changes.</p><p><img alt="" height="282" src="../../engineering/screenshot-2019-07-30-at-09.34.29.png" width="288"></p><p>The ratio <span class="math-tex">\(T_B\over T_C\)</span> is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2</span></label> </p><p><label class="radio"> <input type="radio"> <span>1</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>4</span></label> </p><p><label class="radio"> <input type="radio"> <span>3</span></label> </p></div><div class="q-explanation"><p>Note that <span class="math-tex">\(T_B=T_A\)</span> as these changes occur on isotherms.</p><p>Using the ideal gas law, <span class="math-tex">\(pV=nRT\)</span>:</p><p><span class="math-tex">\({T_A\over T_C}={P_A\over P_C}={400\over 100}\)</span></p><p>HINT: It can be helpful to write <span class="math-tex">\(pVT\)</span> values in pencil at each location on the graph as this will make it easier to spot the equations you can use.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph shows an adiabatic expansion. The amount of gas is such that <span class="math-tex">\(nR =0.01\)</span> Pa m<sup>3 </sup>K<sup>-1</sup>.</p><p><img alt="" height="286" src="../../engineering/screenshot-2019-07-30-at-10.29.50.png" width="363"></p><p>The work done from A to B is 24 J. What is the temperature at B?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>1600 K</span></label> </p><p><label class="radio"> <input type="radio"> <span>1000 K</span></label> </p><p><label class="radio"> <input type="radio"> <span>500 K</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1400 K</span></label> </p></div><div class="q-explanation"><p>An adiabatic change means <span class="math-tex">\(Q = 0 \Rightarrow ΔU = - W\)</span>:</p><p><span class="math-tex">\({3\over 2}nRΔT = -24\)</span></p><p><span class="math-tex">\(ΔT = -1600 \text{ K}\Rightarrow T_B=3000-1600=1400\text{ K}\)</span></p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph shows an adiabatic expansion. The amount of gas is such that <span class="math-tex">\(nR =0.01\)</span> Pa m<sup>3 </sup>K<sup>-1</sup>.</p><p><img alt="" height="286" src="../../engineering/screenshot-2019-07-30-at-10.29.50.png" width="363"></p><p>The work done from A to B is 24 J. How much heat is added?</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>24 J</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>0 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>8 J</span></label> </p><p><label class="radio"> <input type="radio"> <span>-24 J</span></label> </p></div><div class="q-explanation"><p>An adiabatic process means that no heat is exchanged (in or out).</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div><div class="exercise shadow-bottom"><div class="q-question"><p>This <span class="math-tex">\(pV\)</span> graph shows an adiabatic expansion.</p><p><img alt="" height="308" src="../../engineering/screenshot-2019-07-30-at-10.29.50.png" width="391"></p><p>The ratio <span class="math-tex">\(p_AV_A^{5\over 3}\over p_BV_B^{5\over 3}\)</span> is:</p></div><div class="q-answer"><p><label class="radio"> <input type="radio"> <span>2</span></label> </p><p><label class="radio"> <input type="radio"> <span>4</span></label> </p><p><label class="radio"> <input type="radio"> <span>3</span></label> </p><p><label class="radio"> <input class="c" type="radio"> <span>1</span></label> </p></div><div class="q-explanation"><p><span class="math-tex">\(pV^{5\over 3}\)</span> is constant for adiabatic processes.</p><p>HINT: This must be used instead of <span class="math-tex">\({p_AV_A\over T_A}={p_BV_B\over T_B}\)</span> when processes are adiabatic.</p></div><div class="slide-q-actions"><button class="btn btn-default btn-sm btn-xs-block text-xs-center check"><i class="fa fa-check-square-o"></i> Check</button></div></div> </div> </div> <div class="modal-footer slide-quiz-actions"> <div class=""> <div class="pull-left pull-xs-none mb-xs-3"> <button class="btn btn-default d-xs-none btn-prev"> <i class="fa fa-arrow-left"></i> Prev </button> </div> <div 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