Question

Method 2 of DeltaU Work done by a gas in a given process is -20J. Heat given to the gas is 60J. Find change in internal energy of the gas.

Answer 1

change in internal energy of the gas is 80 J

given, work done by a gas in a given process, W = -20 J

Heat given to the gas , ∆Q = 60 J

we have to find change in internal energy of the gas.

from first law of thermodynamics,

∆Q = ∆U + W

⇒60 J = ∆U + (-20 J)

⇒60 J = ∆U - 20J

⇒∆U = 60J + 20J = 80 J

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