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(A) From the given $P-V$ diagram,process $AB$ is an isochoric process (volume is constant).For process $AB$:$\Delta W_{AB} = 0$Using the first law of

Vedclass · Accepted Answer

$1590$

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(A) From the given $P-V$ diagram,process $AB$ is an isochoric process (volume is constant).For process $AB$:$\Delta W_{AB} = 0$Using the first law of thermodynamics,$\Delta Q_{AB} = \Delta U_{AB} + \Delta W_{AB}$.Given $\Delta Q_{AB} = 50 \ J$,so $\Delta U_{AB} = 50 \ J$.Since $U_A = 1500 \ J$,then $U_B = U_A + \Delta U_{AB} = 1500 + 50 = 1550 \ J$.For process $BC$:Given $\Delta Q_{BC} = 0$ (adiabatic process).Work done on the gas is $\Delta W_{BC} = -40 \ J$ (since work is done on the gas,it is negative).Using the first law,$\Delta Q_{BC} = \Delta U_{BC} + \Delta W_{BC}$.$0 = \Delta U_{BC} - 40 \ J \implies \Delta U_{BC} = 40 \ J$.Therefore,$U_C = U_B + \Delta U_{BC} = 1550 + 40 = 1590 \ J$.

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