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Question

The magnitudes of enthalpy changes for irreversible adiabatic expansion of a gas from 1L to 2L is $$\Delta H_1$$ and for reversible adiabatic expansion for the same expansion is $$\Delta H_2$$ then:


A
ΔH1 > ΔH2
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B
ΔH1 < ΔH2
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C
ΔH1 > ΔH2, enthalpy being a state function (ΔH1 = ΔH2)
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D
ΔH1 = ΔH1 & ΔH2 = ΔH2 where ΔH1 & ΔH2 are magnitudes of change in internal energy of gas in these expansions respectively.
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Solution

The correct option is B $$\Delta H_1$$ < $$\Delta H_2$$
Solution:- (B) $$\Delta{{H}_{1}} < \Delta{{H}_{2}}$$
As we know that,
$$\Delta{H} = \Delta{U} + P \Delta{V}$$
For adiabatic process,
$$\Delta{H}  = \cfrac{f}{2} nR \Delta{T} + nR \Delta{T} = n{C}_{P} \Delta{T}$$
$$\therefore \Delta{H} \propto \Delta{T} ..... \left( 1 \right)$$
Now the work done by the gas is maximum in reversible adiabatic process than irreversible adiabatic process.
Now,
$$P \Delta{V} = W$$ 
$$\Rightarrow \Delta{T} \propto W ..... \left( 2 \right)$$
From $${eq}^{n} \left( 1 \right) \& \left( 2 \right)$$, we have
$$\Delta{{H}_{1}} < \Delta{{H}_{2}}$$

Chemistry

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