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(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Step $1$: Calculate the work done during the compression process.Wor

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increases by $250\,J$

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(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Step $1$: Calculate the work done during the compression process.Work done $\Delta W = P \Delta V = 100 	imes (1 - 2) = -100\,J$.Step $2$: Calculate the change in internal energy during the heating process.Given $\Delta Q = 150\,J$ and since it is a constant volume process,$\Delta W = 0$.Therefore,$\Delta Q = \Delta U = 150\,J$.Step $3$: Total change in internal energy.The total change in internal energy is the sum of the changes in both processes. However,the question asks for the net result of the entire sequence.Using $\Delta Q_{total} = \Delta U_{total} + \Delta W_{total}$,where $\Delta Q_{total} = 150\,J$ and $\Delta W_{total} = -100\,J$.$150 = \Delta U + (-100)$.$\Delta U = 150 + 100 = 250\,J$.Thus,the internal energy of the gas increases by $250\,J$.

$A$ gas is compressed from a volume of $2\,m^3$ to a volume of $1\,m^3$ at a constant pressure of $100\,N/m^2$. Then it is heated at constant volume by supplying $150\,J$ of energy. As a result,the internal energy of the gas

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