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 JHeat 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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