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First Law of Thermodynamics Physics Thermodynamics Physics Class 11th

A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure. It absorbs, 40 J of heat during the part AB, no heat during BC and rejects 60 J of heat during CA. A work of 50 J is done on the during the part BC. The internal energy of the gas at A is 1560 J. The work done by the gas during the part CA is:

Option 1 -

20 J

Option 2 -

30 J

Option 3 -

-30 J

Option 4 -

-60J

4 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

Correct Option - 2

Detailed Solution:

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low

...more

Given

Q_AB = +40J

Q_BC = 0J

Q_CA = -60J

W_CA =?

W_BC = -50J (on the gas)

U_A = 1560J

1^st Law on path A to B

W_AB + $Δ U = Q$

Þ 0 + $[U_{B} - U_{A}] = 40$

Þ U_B = U_A + 40

= 1560 + 40

U_B = 1600 J

1^st Low on path B to C

Q_BC = $Δ u + w_{B C}$

Þ 0 = U_c – U_D – 50

U_C = U_B + 50

= 1600 + 50

= 1650 J

U_A – U_C = $Δ U$ (Path C to A) = 1560-1650

= -90 J

1st Low for path C to A

Q_CA = $Δ U_{C A} + w_{C A} + w_{C A}$

Þ -60 = -90 + w_CA

W_CA = +30J

less

Given                                                                           QAB = +40JQBC = 0JQCA = -60JWCA =? WBC = -50J (on the gas)UA = 1560J1st Law on path A to BWAB + <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1814361752"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mi>Δ</mi> <mi>U</mi> <mo>=</mo> <mi>Q</mi> </mrow> </math> Þ 0 + <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1814361753"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mrow> <mo> [</mo> <mrow> <msub> <mrow> <mi>U</mi> </mrow> <mrow> <mi>B</mi> </mrow> </msub> <mo>−</mo> <msub> <mrow> <mi>U</mi> </mrow> <mrow> <mi>A</mi> </mrow> </msub> </mrow> <mo>]</mo> </mrow> <mo>=</mo> <mn>4</mn> <mn>0</mn> </mrow> </math> Þ UB = UA + 40= 1560 + 40UB = 1600 J 1st Low on path B to CQBC = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1814361763"> </o:OLEObject></xml><! [endif]->  <math> <mrow> <mi>Δ</mi> <mi>u</mi> <mo>+</mo> <msub> <mrow> <mi>w</mi> </mrow> <mrow> <mi>B</mi> <mi>C</mi> </mrow> </msub> </mrow> </math> Þ 0 = Uc – UD – 50UC = UB + 50= 1600 + 50= 1650 JUA – UC = <math> <mrow> <mi>Δ</mi> <mi>U</mi> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1814361764"> </o:OLEObject></xml><! [endif]->  (Path C to A) = 1560-1650= -90 J1st Low for path C to AQCA = <math> <mrow> <mi>Δ</mi> <msub> <mrow> <mi>U</mi> </mrow> <mrow> <mi>C</mi> <mi>A</mi> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>w</mi> </mrow> <mrow> <mi>C</mi> <mi>A</mi> </mrow> </msub> <mo>+</mo> <msub> <mrow> <mi>w</mi> </mrow> <mrow> <mi>C</mi> <mi>A</mi> </mrow> </msub> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1814361765"> </o:OLEObject></xml><! [endif]-> Þ -60 = -90 + wCAWCA = +30J

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