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A sample of argon of $1 \mathrm{~atm}$ pressure and $300 \mathrm{~K}$ expands reversibly and adiabatically from $1.25 \mathrm{dm}^3$ to $2.5 \mathrm{dm}^3$. Calculate the approximate enthalpy (in J) change
(i) $C_V$ for argon is $12.48 \mathrm{JK}^{-1}$
(ii) Assume argon to be an ideal gas
(iii) $\Delta T=111.5 \mathrm{~K}$
Options:
(i) $C_V$ for argon is $12.48 \mathrm{JK}^{-1}$
(ii) Assume argon to be an ideal gas
(iii) $\Delta T=111.5 \mathrm{~K}$
Solution:
2214 Upvotes
Verified Answer
The correct answer is:
117
We know that, $\Delta H=n C_p \Delta T$
Number of moles of argon $(n)=\frac{p V}{R T}$
Given, $p=1 \mathrm{~atm}, \quad V=1.25 \mathrm{dm}^3, \quad T=300 \mathrm{~K}$
$\therefore n=\frac{1 \times 1.25}{0.082 \times 300}=0.05 \mathrm{~mol}$
Also, $C_p C_V=12.48 \mathrm{JK}^{-1}$
$$
C_p=C_V+R=1248+8.314=20.794
$$
On substituting the above values in Eq. (i), we get
$$
\Delta H=0.05 \times 20.794 \times 111.5=115.92 \approx 117 \mathrm{~J}
$$
Number of moles of argon $(n)=\frac{p V}{R T}$
Given, $p=1 \mathrm{~atm}, \quad V=1.25 \mathrm{dm}^3, \quad T=300 \mathrm{~K}$
$\therefore n=\frac{1 \times 1.25}{0.082 \times 300}=0.05 \mathrm{~mol}$
Also, $C_p C_V=12.48 \mathrm{JK}^{-1}$
$$
C_p=C_V+R=1248+8.314=20.794
$$
On substituting the above values in Eq. (i), we get
$$
\Delta H=0.05 \times 20.794 \times 111.5=115.92 \approx 117 \mathrm{~J}
$$
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