A perfect gas goes from state A to another st | vedclass

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(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.In the first process:$\Delta Q_1 = 8 	imes 10^5 \ J$ and $\Delta W_

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Work done on the gas is $0.5 	imes 10^5 \ J$

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(A) According to the first law of thermodynamics,$\Delta Q = \Delta U + \Delta W$.In the first process:$\Delta Q_1 = 8 	imes 10^5 \ J$ and $\Delta W_1 = 6.5 	imes 10^5 \ J$.Therefore,$\Delta U = \Delta Q_1 - \Delta W_1 = (8 - 6.5) 	imes 10^5 \ J = 1.5 	imes 10^5 \ J$.Since the initial and final states are the same for both processes,the change in internal energy $\Delta U$ remains constant.In the second process:$\Delta Q_2 = 10^5 \ J$ and $\Delta U = 1.5 	imes 10^5 \ J$.Using $\Delta Q_2 = \Delta U + \Delta W_2$:$10^5 = 1.5 	imes 10^5 + \Delta W_2$.$\Delta W_2 = 10^5 - 1.5 	imes 10^5 = -0.5 	imes 10^5 \ J$.$A$ negative value for work done by the gas indicates that work is done on the gas. Thus,the work done on the gas is $0.5 	imes 10^5 \ J$.

$A$ 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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