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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Answer:
Correct Answer: 71. $(-116.4$ $ J)$
Solution:
Given : $C _V=12.49 \Rightarrow C _p=20.8$
$\Rightarrow \quad \dfrac{C _p}{C _V}=\gamma=1.66$
In case of reversible adiabatic expansion :
$ \begin{aligned} & T V^{\gamma-1}=\text { constant } \\ \Rightarrow \quad \dfrac{T _2}{T _1} & =\left(\dfrac{V _1}{V _2}\right)^{\gamma-1}=\left(\dfrac{V _1}{V _2}\right)^{0.66} \\ \Rightarrow \quad T _2 & =T _1\left(\dfrac{V _1}{V _2}\right)^{0.66} \\ & =300\left(\dfrac{1}{2}\right)^{0.66}=189.86 K \\ \Rightarrow \quad \Delta H & =n C _p \Delta T \quad = \dfrac{1 \times 1.25}{0.082 \times 300} \times 20.8 \times(189.86-300) J \\ & =-116.4 J \end{aligned} $
BETA
