JEE MAIN - Chemistry (2017 - 9th April Morning Slot - No. 19)
A gas undergoes change from state A to state B. In this process, the heat absorbed
and work done by the gas is 5 J and 8 J, respectively. Now gas is brought back to
A by another process during which 3 J of heat is evolved. In this reverse process of
B to A :
10 J of the work will be done by the gas.
6 J of the work will be done by the gas.
10 J of the work will be done by the surrounding on gas.
6 J of the work will be done by the surrounding on gas.
Explanation
From first law of thermodynamics, we have
$$\Delta$$U = q + w
$$\bullet$$ For state A to B : q = +5 J, w = $$-$$8 J and
$$\Delta$$UAB = 5 + ($$-$$8) = $$-$$3 J
$$\bullet$$ For state B to A : q = $$-$$3 J
Since the internal energy is a state function, we have $$\Delta$$UBA = $$-$$ $$\Delta$$UAB. Therefore,
$$\Delta$$UBA = q + w
3 = $$-$$3 + w $$\Rightarrow$$ w = 6 J
Positive value of work indicates that 6 J of work is done by the surrounding on gas.
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