A thermodynamic system is taken through the cycle ABCD as shown in figure. Heat rejected by the gas during the cycle is

 

Correct Option - 1


Detailed Solution:

In a cyclic process,   $\Delta U=0.$

In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise 

$\Delta W=$ Area of rectangle $ABCD=-P\left(2V\right)$

$=-2PV$

According to first law of thermodynamics

$\Delta Q=\Delta U+\Delta W$ or $\Delta Q=\Delta W$ $\left(As\Delta U=0\right)$

i.e., heat supplied to the system is equal to the work done

So heat absorbed, $\Delta Q=\Delta W=-2PV$

Heat rejected by the gas $=2PV$

<p>In a cyclic process, &nbsp;<span contenteditable="false"> <math> <mrow> <mi>Δ</mi> <mi>U</mi> <mo>=</mo> <mn>0</mn> <mo>.</mo> </mrow> </math> </span>&lt;!- [if gte mso 9]>&lt;xml&gt; <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1819715275"> </o:OLEObject>&lt;/xml&gt;&lt;! [endif]-&gt;</p><p>In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise&nbsp;</p><div><div><picture><source srcset="" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/1758189795phpwfNita.jpeg" alt="" width="149" height="137" loading="lazy"></picture></div><div><span contenteditable="false"> <math> <mrow> <mi>Δ</mi> <mi>W</mi> <mo>=</mo> </mrow> </math> </span>Area of rectangle <span contenteditable="false"> <math> <mrow> <mi>A</mi> <mi>B</mi> <mi>C</mi> <mi>D</mi> <mo>=</mo> <mo>−</mo> <mi>P</mi> <mrow> <mo> (</mo> <mrow> <mn>2</mn> <mi>V</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> </span>&lt;!- [if gte mso 9]>&lt;xml&gt; <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1819715381"> </o:OLEObject>&lt;/xml&gt;&lt;! [endif]-&gt;</div><div><span contenteditable="false"> <math> <mrow> <mo>=</mo> <mo>−</mo> <mn>2</mn> <mi>P</mi> <mi>V</mi> </mrow> </math> </span></div><div><p>According to first law of thermodynamics</p><p><span contenteditable="false"> <math> <mrow> <mi>Δ</mi> <mi>Q</mi> <mo>=</mo> <mi>Δ</mi> <mi>U</mi> <mo>+</mo> <mi>Δ</mi> <mi>W</mi> </mrow> </math> </span> or <span contenteditable="false"> <math> <mrow> <mi>Δ</mi> <mi>Q</mi> <mo>=</mo> <mi>Δ</mi> <mi>W</mi> </mrow> </math> </span> <span contenteditable="false"> <math> <mrow> <mrow> <mo> (</mo> <mrow> <mi>A</mi> <mi>s</mi> <mi>Δ</mi> <mi>U</mi> <mo>=</mo> <mn>0</mn> </mrow> <mo>)</mo> </mrow> </mrow> </math> </span></p></div></div><p><em>i</em>.<em>e</em>., heat supplied to the system is equal to the work done</p><p>So heat absorbed, <span contenteditable="false"> <math> <mrow> <mi>Δ</mi> <mi>Q</mi> <mo>=</mo> <mi>Δ</mi> <mi>W</mi> <mo>=</mo> <mo>−</mo> <mn>2</mn> <mi>P</mi> <mi>V</mi> </mrow> </math> </span></p><p>Heat rejected by the gas <span contenteditable="false"> <math> <mrow> <mo>=</mo> <mn>2</mn> <mi>P</mi> <mi>V</mi> </mrow> </math> </span></p>

0 Upvotes 0 Upvotes 0 Downvotes

• • Report Abuse
• Comment