Thermodynamics and Thermochemistry 1 Question 68
71. A sample of argon gas at $1 \mathrm{~atm}$ pressure and $27^{\circ} \mathrm{C}$ expands reversibly and adiabatically from $1.25 \mathrm{dm}^{3}$ to $2.50 \mathrm{dm}^{3}$. Calculate the enthalpy change in this process $C_{V_{m}}$ for argon is $12.49 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$.
(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 :
$$ \begin{aligned} & T V^{\gamma-1}=\text { constant } \ \Rightarrow \quad \frac{T_{2}}{T_{1}} & =\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=\left(\frac{V_{1}}{V_{2}}\right)^{0.66} \ \Rightarrow \quad T_{2} & =T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{0.66} \ & =300\left(\frac{1}{2}\right)^{0.66}=189.86 \mathrm{~K} \ \Rightarrow \quad \Delta H & =n C_{p} \Delta T \quad \frac{1 \times 1.25}{0.082 \times 300} \times 20.8 \times(189.86-300) \mathrm{J} \ & =-116.4 \mathrm{~J} \end{aligned} $$
