<p>Consider the figure below.</p> <div class=\"figure\"><img src=\"http://www.we

Question

<p>Consider the figure below.</p>
<div class="figure"><img src="http://www.webassign.net/walker/18-23alt.gif" border="0" alt="" /></div>
<div class="indent">(a) Find the work done by a monatomic ideal gas as it expands from point A to point C along the path shown in the figure. (The horizontal axis is marked in increments of <span>1</span> m³.) <br /><span class="qTextField"> <span class="box_num" title="Points: /1">1</span><input id="RN_1130354_3_0_1135749" dir="ltr" name="RN_1130354_3_0_1135749" size="10" type="text" /></span> MJ<br /><br /> (b) If the temperature of the gas is <span>235</span> K at point A, what is its temperature at point C?<br /><span class="qTextField"> <span class="box_num" title="Points: /1">2</span><input id="RN_1130354_3_1_1135749" dir="ltr" name="RN_1130354_3_1_1135749" size="10" type="text" /></span> K<br /><br /> (c) How much heat has been added to or removed from the gas during this process?<br /> <span class="qTextField"> <span class="box_num" title="Points: /1">3</span><input id="RN_1130354_3_2_1135749" dir="ltr" name="RN_1130354_3_2_1135749" size="10" type="text" /></span> MJ</div>
<p>&#160;</p>

Explanation / Answer

        (a) Work done by the gas is area under the curve             The number of squares under the curve are 6.5 squares           The work done by the gas is     = 6.5 ( 2.0*105 Pa) ( 1m3 )   = 15*105   J       (b) Initial temperature at a point A is   TA   = 235 K          initial presure PA   =    2.0*105 Pa , initial volume at A is   VA = 1m3           Pressure at point C is PC =  2.0*105 Pa  , Volume at point C is     VC = 5m3           From ideal gas equation              PA VA /   TA      =   PC    VC /   TC           Then solving for TC    =    PC    VC /       (  PA VA /   TA  )               plug all values we get      TC    =   1175 K      (c) According   to first law of thermodynamics    Heat   added   to the system     Q = dU + W               here   Change in internal energy   dU   = nCv dT              For monoatomis gas   n =1   and Cv   = (3/2)R           then  dU   = (3/2)R    dT        =   (3/2) P dV             dU   = (3/2) (2.0*105 Pa) ( 5m3   -    1m3 )    =   12*105   J     Then Q =    12*105   J + 15*105   J     = 27 *105   J               Then solving for TC    =    PC    VC /       (  PA VA /   TA  )               plug all values we get      TC    =   1175 K      (c) According   to first law of thermodynamics    Heat   added   to the system     Q = dU + W               here   Change in internal energy   dU   = nCv dT              For monoatomis gas   n =1   and Cv   = (3/2)R           then  dU   = (3/2)R    dT        =   (3/2) P dV             dU   = (3/2) (2.0*105 Pa) ( 5m3   -    1m3 )    =   12*105   J     Then Q =    12*105   J + 15*105   J     = 27 *105   J    

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