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

Suggest Corrections  0  Similar questions
View More  People also searched for
View More 