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Standard XII

Chemistry

Thermodynamic Processes

A thermodynam...

Question

A thermodynamical process is shown in the figure with $P_{A} = 3 \times P_{a t m}, V_{A} = 2 \times 10^{- 4} m^{3}, P_{B} = 8 \times P_{a t m}, V_{C} = 5 \times 10^{- 4} m^{3} .$ In the process $A B$ and $B C$ , $600 J$ and $200 J$ heat are added to the system. Find the change in internal energy of the system in the process $C A$ . $[1 P_{a t m} = 10^{5} N / m^{2}]$

560 J

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- 560 J

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- 240 J

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+ 240 J

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Solution

The correct option is B $- 560 J$
for process AB which is a constant volume process so work done is zero.
$U_{A B} = Q_{A B} = 600 J$
for process BC work done will be

$W_{B C}$ = $8 \times 10^{5} [5 \times 10^{- 4} - 2 \times 10^{- 4}] = 240 J$
$U_{B C}$ = $Q_{B C} - W_{B C}$
$U_{B C} = 200 - 240 = - 40 J$

now we know that internal energy for a cyclic process will be zero
hence,we can write $U_{A B}$ + $U_{B C}$ + $U_{C A} = 0$
$U_{C A} = - 600 + 40 = - 560 J$

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A thermodynamical process is shown in the figure with PA=3×Patm,VA=2×10−4m3,PB=8×Patm,VC=5×10−4m3. In the process AB and BC, 600J and 200J heat are added to the system. Find the change in internal energy of the system in the process CA. [1Patm=105N/m2]