A thermodynamic process is shown in the figur | vedclass

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(A) The process $AB$ is an isochoric process (constant volume), so no work is done during $AB$ $(W_{AB} = 0)$.The process $BC$ is an isobaric process

Vedclass · Accepted Answer

$560$

Vedclass · Answer

(A) The process $AB$ is an isochoric process (constant volume), so no work is done during $AB$ $(W_{AB} = 0)$.The process $BC$ is an isobaric process (constant pressure). Therefore, the work done during process $BC$ is:$W_{BC} = P_B 	imes (V_D - V_A) = 8 	imes 10^4 	imes (5 	imes 10^{-3} - 2 	imes 10^{-3}) = 8 	imes 10^4 	imes 3 	imes 10^{-3} = 240 	ext{ J}$.The total heat added to the system to take it from state $A$ to state $C$ via path $ABC$ is:$\Delta Q = Q_{AB} + Q_{BC} = 600 + 200 = 800 	ext{ J}$.According to the first law of thermodynamics, $\Delta Q = \Delta E_{int} + \Delta W$.Since internal energy is a state function, the change in internal energy $\Delta E_{int}$ for path $AC$ is the same as for the path $ABC$.$\Delta E_{int} = \Delta Q - \Delta W = 800 - 240 = 560 	ext{ J}$.

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