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First Law of Thermodynamics *** A non-ideal gas goes through the cycle ABCA (shown in the figure below). If 7950 J of heat is
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Answer #1

for the total work done during the process

W = area inside the triangle

W = 0.50 * 1 *10^5 * 2 *10^-2

W = 1000 J

Now, as during the cycle

the change in internal energy is zero

hence , Qcycle = W

7950 - 1750 + QCA = 1000

QCA = 5200 J

the heat added during CA is 5200 J

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