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Heat and Thermodynamics question

2019 · 9 Jan · Shift 1 · Q50
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Heat and Thermodynamics question

2019 · 9 Jan · Shift 1 · Q50

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A gas can be taken from A to B via two different processes ACB and ADB. JEE Main 2019 (Online) 9th January Morning Slot Physics - Heat and Thermodynamics Question 364 English When path ACB is used 60 J of heat flows into the system and 30 J of work is done by the system. If path ADB is used work done by the system is 10 J. The heat Flow into the system in path ADB is :
  1. A
    40 J
  2. B
    80 J
  3. C
    100 J
  4. D
    20 J
View written solutionFree

Correct answer: A

  1. Use the first law of thermodynamics:

ΔU=Q−W\Delta U = Q - WΔU=Q−W

where:

  • QQQ = heat added to the system
  • WWW = work done by the system
  • ΔU\Delta UΔU = change in internal energy
  1. Since the gas goes from the same initial state AAA to the same final state BBB, the change in internal energy is the same for both paths.

For path ACBACBACB

Given:

  • Heat added, QACB=60 JQ_{ACB} = 60\,\text{J}QACB​=60J
  • Work done, WACB=30 JW_{ACB} = 30\,\text{J}WACB​=30J

So, ΔU=QACB−WACB=60−30=30 J\Delta U = Q_{ACB} - W_{ACB} = 60 - 30 = 30\,\text{J}ΔU=QACB​−WACB​=60−30=30J

  1. Now for path ADBADBADB: Given:
  • Work done, WADB=10 JW_{ADB} = 10\,\text{J}WADB​=10J
  • Same ΔU=30 J\Delta U = 30\,\text{J}ΔU=30J

Using ΔU=QADB−WADB\Delta U = Q_{ADB} - W_{ADB}ΔU=QADB​−WADB​

30=QADB−1030 = Q_{ADB} - 1030=QADB​−10

QADB=40 JQ_{ADB} = 40\,\text{J}QADB​=40J

  1. Therefore, the heat flowing into the system along path ADBADBADB is:

40 J\boxed{40\,\text{J}}40J​

  1. Option check:
  • A: 40 J40\,\text{J}40J ✅
  • B: 80 J80\,\text{J}80J ❌
  • C: 100 J100\,\text{J}100J ❌
  • D: 20 J20\,\text{J}20J ❌
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