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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?

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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.

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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! ;)


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

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – jonsca Feb 25 '18 at 2:53

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