A perfect gas goes from state $A$ to another state $B$ by absorbing $8 \times {10^5}J$ of heat and doing $6.5 \times {10^5}J$ of external work. It is now transferred between the same two states in another process in which it absorbs ${10^5}J$ of heat. Then in the second process
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(a) In first process using $\Delta Q = \Delta U + \Delta W$
==> $8 \times {10^5} = \Delta U + 6.5 \times {10^5}$==> $\Delta U = 1.5 \times 10J$
Since final and initial states are same in both process
So $\Delta U$ will be same in both process
For second process using $\Delta Q = \Delta U + \Delta W$
==> ${10^5} = 1.5 \times {10^5} + \Delta W$==>$\Delta W = - 0.5 \times {10^5}J$
==> $8 \times {10^5} = \Delta U + 6.5 \times {10^5}$==> $\Delta U = 1.5 \times 10J$
Since final and initial states are same in both process
So $\Delta U$ will be same in both process
For second process using $\Delta Q = \Delta U + \Delta W$
==> ${10^5} = 1.5 \times {10^5} + \Delta W$==>$\Delta W = - 0.5 \times {10^5}J$
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