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प्रश्न
A gas performs 0.320 kJ work on surrounding and absorbs 120 J of heat from the surrounding. Hence, change in internal energy is ______.
पर्याय
200 J
120.32 J
- 200 J
440 J
MCQ
रिकाम्या जागा भरा
उत्तर
A gas performs 0.320 kJ work on surrounding and absorbs 120 J of heat from the surrounding. Hence, change in internal energy is - 200 J.
Explanation:
According to first law of thermodynamics,
Δ U = q + W
Since, work is done on the surrounding.
So, W = - 0.320 kJ
= - 320 J
q = 120 J
Δ U = - 320 + 120
= - 200 J
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First Law of Thermodynamics
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