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(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Here,the work done on the gas during compression is $\Delta W = P \D

Vedclass · Accepted Answer

$400$

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(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Here,the work done on the gas during compression is $\Delta W = P \Delta V$.Since the volume changes from $10 \ m^{3}$ to $4 \ m^{3}$,$\Delta V = (4 - 10) \ m^{3} = -6 \ m^{3}$.Thus,$\Delta W = 50 	imes (-6) = -300 \ J$.The heat supplied is $\Delta Q = 100 \ J$.Using the first law: $\Delta U = \Delta Q - \Delta W$.$\Delta U = 100 - (-300) = 100 + 300 = 400 \ J$.

$A$ gas is compressed at a constant pressure of $50 \ N/m^{2}$ from a volume of $10 \ m^{3}$ to $4 \ m^{3}$. Then,the gas is heated by supplying $100 \ J$ of energy. The increase in its internal energy will be ..... $J$.

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