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Thermodynamics physics ncert solutions class 11th Physics Class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Twelve

A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig.. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R)

AB : constant volume

BC : constant pressure

CD : adiabatic

DA : constant pressure

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Payal Gupta | Contributor-Level 10

4 months ago

This is a long answer type question as classified in NCERT Exemplar

(a) For process AB

Volume is constant , hence work done dW=0

dQ=dU+dW=dU+0=dU

= nC_vdT= nC_v(T_B-T_A)

= $\frac{3}{2} R (T_{B} - T_{A})$

= $\frac{3}{2} (R T_{B} - R T_{A}) = \frac{3}{2} (P_{B} V_{B} - P_{A} V_{A})$

Heat exchanged = $\frac{3}{2} (P_{B} V_{B} - P_{A} V_{A})$

(b) For process BC , p =constant

dQ= dU+dW = $\frac{3}{2} R (T_{C} - T_{B}) + P_{B} (V_{C} - V_{B})$

heat exchanged = $\frac{5}{2} P_{B} (V_{C} {- V}_{A})$

(c) For process CD , because CD is adiabatic , dQ= heat exchanged =0

(d) DA involves compression of gas from V_D to V_A at constant pressure P_A

heat transferred as similar way as BC₁

hence dQ = $\frac{5}{2}$ P_A(V_A-V_D)

This is a long answer type question as classified in NCERT Exemplar(a) For process ABVolume is constant , hence work done dW=0dQ=dU+dW=dU+0=dU = nCvdT= nCv(TB-TA) = <math> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>R</mi> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </math> = <math> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mfenced separators="|"> <mrow> <mrow> <mi>R</mi> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </msub> <mo>-</mo> <mi>R</mi> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> <mo>=</mo> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </math> Heat exchanged = <math> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> </math> (b) For process BC , p =constantdQ= dU+dW  = <math> <mi> </mi> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>R</mi> <mfenced separators="|"> <mrow> <mrow> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>C</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfenced> <mo>+</mo> <msub> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </msub> <mo>(</mo> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>C</mi> </mrow> </mrow> </msub> <mo>-</mo> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </msub> <mo>)</mo> </math> heat exchanged = <math> <mfrac> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <msub> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> <mrow> <mrow> <mi>B</mi> </mrow> </mrow> </msub> <mo>(</mo> <msub> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>C</mi> </mrow> </mrow> </msub> <msub> <mrow> <mrow> <mo>-</mo> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> <mo>)</mo> </math> (c) For process CD , because CD is adiabatic , dQ= heat exchanged =0(d) DA involves compression of gas from VD to VA at constant pressure PAheat transferred as similar way as BC1hence dQ = <math> <mfrac> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> PA(VA-VD)

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A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig.. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R)

AB : constant volume

BC : constant pressure

CD : adiabatic

DA : constant pressure

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A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig.. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R) AB : constant volume BC : constant pressure CD : adiabatic DA : constant pressure

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A cycle followed by an engine (made of one mole of an ideal gas in a cylinder with a piston) is shown in Fig.. Find heat exchanged by the engine, with the surroundings for each section of the cycle. (Cv = (3/2) R)

AB : constant volume

BC : constant pressure

CD : adiabatic

DA : constant pressure