Question

Method 2 of Q In a given process work done on a gas is 40 J and increases in its internal energy is 10J. Find heat given or taken to/from the gas in this process.

Answer 1

30J heat is taken out from the gas in this process

given, in a given process work done on a gas, W = 40J

so, workdone by the gas , W = -40 J

increases in its internal energy, ∆U = 10J

we have to find heat taken to/from the has in this process.

from First law of thermodynamics,

∆Q = ∆U + W

⇒∆Q = -40 J + 10J = -30 J

here negative sign indicates that heat is taken out from the gas in this process.

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Answer 2

Heat taken from the gas is -30J.

As we know that change in internal energy=heat + work done.

Hence heat taken from the gas is -30