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Question

A perfect gas goes from a state A to state B by absorbing 8×105 J and by doing 6.5×105 J of external work. It is taken from same initial state A to final state B in another process in which it absorbs 105 J of heat, then work done in the second process

A
on gas is 105 J
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B
on gas is 0.5×105 J
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C
by gas is 105 J
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D
by gas is 0.5×105 J
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Solution

The correct option is B on gas is 0.5×105 J
Heat is absorbed by system {gas}, ΔQ=+8×105 J
work done by system, ΔW=+6.5×105 J
According to the 1st law of thermodynamics
ΔQ=ΔU+ΔW
8×105=ΔU+6.5×105
ΔU=+1.5×105J ........(1)
For the second process,
Heat absorbed by the system ΔQ=+105 J
Again using First law,ΔQ=ΔU+ΔW
We write that,
ΔQ=ΔU+ΔW
ΔU will stay the same as initial and final states of gas is same.
Now, From the data given in the question and (1)
105=1.5×105+ΔW
ΔW=0.5×105 J
(negative sign indicates work is being done on the gas).
Thus, option (b) is the correct answer.

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