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 - <p>20 J</p>
Option 2 - <p>30 J</p>
Option 3 - <p>-30 J</p>
Option 4 - <p>&nbsp;-60J</p>
13 Views|Posted 6 months ago
Asked by Shiksha User
1 Answer
A
6 months ago
Correct Option - 2
Detailed Solution:

Given                                                                           

QAB = +40J

QBC = 0J

QCA = -60J

WCA =?

WBC = -50J (on the gas)

UA = 1560J

1st Law on path A to B

WAB +   Δ U = Q

Þ 0 +    [ U B U A ] = 4 0

Þ UB = UA + 40

= 1560 + 40

UB = 1600 J

 1st Low on path B to C

QBC =   Δ u + w B C

Þ 0 = Uc – UD – 50

UC = UB + 50

= 1600 +

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Physics Thermodynamics 2025

Physics Thermodynamics 2025

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