Solution:
Using the First Law of Thermodynamics, \(Delta Q = Delta U + W\). Here, \(Delta Q = 10\text{ J}\) and \(Delta U = -5\text{ J}\) (decrease). Substituting these values, \(10 = -5 + W\), which gives \(W = 15\text{ J}\).
If \(10\text{ J}\) of heat energy is supplied to a gas sample and \(5\text{ J}\) of its internal energy decreases during the process, then work done by the gas will be
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