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(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Given:Heat released by the gas,$\Delta Q = -20 \ J$ (heat lost by th

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$30$

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(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.Given:Heat released by the gas,$\Delta Q = -20 \ J$ (heat lost by the system is negative).Work done on the gas,$\Delta W = -10 \ J$ (work done on the system is negative in the convention $\Delta Q = \Delta U + \Delta W$).Initial internal energy,$U_i = 40 \ J$.Substituting the values into the equation:$-20 = (U_f - 40) + (-10)$$-20 = U_f - 40 - 10$$-20 = U_f - 50$$U_f = 50 - 20 = 30 \ J$.Wait,re-evaluating the sign convention: If $\Delta Q = \Delta U + W$ where $W$ is work done $BY$ the gas,then $W = -10 \ J$. Thus,$-20 = \Delta U - 10$,so $\Delta U = -10 \ J$. Since $\Delta U = U_f - U_i$,we have $U_f - 40 = -10$,which gives $U_f = 30 \ J$.

In a thermodynamic process,the pressure of a fixed mass of a gas is changed in such a manner that the gas molecules release $20 \ J$ of heat and $10 \ J$ of work is done on the gas. If the initial internal energy of the gas was $40 \ J$,then the final internal energy will be ........ $J$.

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