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(A) According to the first law of thermodynamics: $\Delta Q = \Delta U + \Delta W$.Here,$\Delta Q = 100\,J$ (heat supplied to the system).The work don

Vedclass · Accepted Answer

$400$

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(A) According to the first law of thermodynamics: $\Delta Q = \Delta U + \Delta W$.Here,$\Delta Q = 100\,J$ (heat supplied to the system).The work done by the gas is $\Delta W = P \Delta V = P(V_f - V_i)$.Given $P = 50\,N/m^2$,$V_i = 10\,m^3$,and $V_f = 4\,m^3$.$\Delta W = 50 	imes (4 - 10) = 50 	imes (-6) = -300\,J$.Substituting these values into the first law equation:$100 = \Delta U + (-300)$.$\Delta U = 100 + 300 = 400\,J$.Therefore,the increase in internal energy is $400\,J$.

The volume of a gas at a constant pressure of $50\,N/m^2$ changes from $10\,m^3$ to $4\,m^3$. If $100\,J$ of heat is supplied to the gas,the increase in internal energy is ....... $J$.

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