please solve this Q 33.
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Option 2 is the right answer... Since there is no change in number of moles of gas, so del S is zero. Therefore,
∆H = ∆U +T∆S
∆H = ∆U +T×0
∆H =∆U
∆H = ∆U +T∆S
∆H = ∆U +T×0
∆H =∆U
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