An ideal gas is taken from state-1 to state-2 | vedclass

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(A) According to the first law of thermodynamics,$\Delta U = Q - W$,which implies $Q - W = \Delta U$.Since internal energy $(U)$ is a state function,t

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$Q_A - W_A = Q_D - W_D$

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(A) According to the first law of thermodynamics,$\Delta U = Q - W$,which implies $Q - W = \Delta U$.Since internal energy $(U)$ is a state function,the change in internal energy $(\Delta U)$ depends only on the initial and final states.For all paths $(A, B, C, D)$,the initial state is $1$ and the final state is $2$.Therefore,the change in internal energy is the same for all paths: $\Delta U_A = \Delta U_B = \Delta U_C = \Delta U_D$.Since $Q - W = \Delta U$,it follows that $Q_A - W_A = Q_B - W_B = Q_C - W_C = Q_D - W_D$.Thus,the expression $Q - W$ is constant for all paths connecting the same two states.Comparing this with the given options,option $A$ is correct because $Q_A - W_A = \Delta U$ and $Q_D - W_D = \Delta U$,so $Q_A - W_A = Q_D - W_D$.

An ideal gas is taken from state-$1$ to state-$2$ through paths $A, B, C$ and $D$ as shown in the $P-V$ diagram. Let $Q, W$ and $\Delta U$ represent the heat supplied,work done,and change in internal energy respectively. Then:

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