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(A) According to the First Law of Thermodynamics $(FLOT)$: $\Delta Q = \Delta U + \Delta W$.Here,$\Delta Q$ is the heat added to the system,$\Delta U$

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Increased by $400 \ J$

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(A) According to the First Law of Thermodynamics $(FLOT)$: $\Delta Q = \Delta U + \Delta W$.Here,$\Delta Q$ is the heat added to the system,$\Delta U$ is the change in internal energy,and $\Delta W$ is the work done by the gas.Given: Pressure $P = 50 \ N/m^2$,initial volume $V_i = 10 \ m^3$,final volume $V_f = 4 \ m^3$,and heat added $\Delta Q = 100 \ J$.The work done by the gas is $\Delta W = P \Delta V = P(V_f - V_i) = 50 	imes (4 - 10) = 50 	imes (-6) = -300 \ J$.Substituting these values into the $FLOT$ equation: $100 = \Delta U + (-300)$.Therefore,$\Delta U = 100 + 300 = 400 \ J$.Since $\Delta U$ is positive,the internal energy increases by $400 \ J$.

$A$ gas is compressed at a constant pressure of $50 \ N/m^2$ from a volume of $10 \ m^3$ to a volume of $4 \ m^3$. Energy of $100 \ J$ is then added to the gas by heating. Its internal energy is:

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