Expansion under isobaric adiabatic condition


> 
> A 3 mole sample of a triatomic gas at $\pu{300 K}$ is allowed to expand under isobaric adiabatic condition from $\pu{5L}$ to $\pu{40L}$. The value of change in enthalpy is: 
> 
> 
> 1. $\pu{12.46 KJ}$
> 2. $\pu{-14.965 KJ}$
> 3. $\pu{-24.62 KJ}$
> 4. $\pu{-10.24 KJ}$
> 
> 
> 


In case of adiabatic process, there is no transfer of heat so $Q=0$. Again for an isobaric process, pressure is constant that is $\Delta P=0$. 
So, change in enthalpy:


$$\Delta H = \Delta U + \Delta(PV) = Q + W + P\Delta V + V\Delta P = W + P\Delta V = -P\Delta V + P\Delta V = 0$$


Thus, we see that change in enthalpy is zero. But the answer matches with none of the options. What is incorrect in my solution?




If, at time = 0, we suddenly drop the external force per unit area to $P\_{\text{ext}}$ and hold it at this value until our adiabatic system equilibriates, the first law of thermodynamics for this situation reads:


$$nC\_\mathrm v(T\_2-T\_1)=-P\_{\text{ext}}(V\_2-V\_1)=-\frac{nRT\_2}{V\_2}(V\_2-V\_1)=-nRT\_2\left[1-\frac{1}{8}\right]$$


Solving this equation for $T\_2$ yields: $$T\_2=\frac{T\_1}{\left[1+\frac{7}{8}\frac{R}{C\_\mathrm v}\right]}$$


But, for $C\_v=3R$, this gives:$$T\_2=0.774T\_1=\pu{232 K}$$


So, $$\Delta H=3(4R)(232-300)=\pu{-6.78 kJ}$$


So the final temperature is higher than in the adiabatic reversible case, and the decrease in enthalpy is less.


 2

**I figured out the problem!**


There is a misprint. The word *isobaric* should be deleted. Without it, the question makes complete sense. Here's how:


Since it is now a *reversible* process, apply $TV^{\gamma-1}=\text{constant}$, with $\gamma=1+\frac2f$ and $f=6$ to get $T\_2=\pu{150K}$. Now, $\Delta H=n\cdot C\_p\cdot\Delta T=3\cdot(1+\frac62)\cdot R\cdot(-150) \pu{J}=-\pu{14.965 KJ}$ as given a correct answer in your book.


Thank me later! ;)




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*Note: This answer has been completely rewritten. The previous version of this answer can be found in the revision history.*


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