Thermodynamics and Thermochemistry - Result Question 71
####71. A sample of argon gas at $1 atm$ pressure and $27^{\circ} C$ expands reversibly and adiabatically from $1.25 dm^{3}$ to $2.50 dm^{3}$. Calculate the enthalpy change in this process $C _{V _m}$ for argon is $12.49 JK^{-1} mol^{-1}$.
(2000, 4M)
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Solution:
- Given : $C _V=12.49 \Rightarrow C _p=20.8$
$\Rightarrow \quad \frac{C _p}{C _V}=\gamma=1.66$
In case of reversible adiabatic expansion :
$$ \begin{aligned} & T V^{\gamma-1}=\text { constant } \ \Rightarrow \quad \frac{T _2}{T _1} & =\left(\frac{V _1}{V _2}\right)^{\gamma-1}=\left(\frac{V _1}{V _2}\right)^{0.66} \ \Rightarrow \quad T _2 & =T _1\left(\frac{V _1}{V _2}\right)^{0.66} \ & =300\left(\frac{1}{2}\right)^{0.66}=189.86 K \ \Rightarrow \quad \Delta H & =n C _p \Delta T \quad \frac{1 \times 1.25}{0.082 \times 300} \times 20.8 \times(189.86-300) J \ & =-116.4 J \end{aligned} $$
