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(D) The total internal energy $U$ of a gas mixture is the sum of the internal energies of its individual components: $U = U_1 + U_2$.For oxygen $(O_2)

Vedclass · Accepted Answer

$11$

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(D) The total internal energy $U$ of a gas mixture is the sum of the internal energies of its individual components: $U = U_1 + U_2$.For oxygen $(O_2)$,which is a diatomic gas,the degrees of freedom $f_1 = 5$ (neglecting vibrational modes). Thus,$C_{v1} = \frac{f_1}{2}R = \frac{5}{2}R$.For argon $(Ar)$,which is a monatomic gas,the degrees of freedom $f_2 = 3$. Thus,$C_{v2} = \frac{f_2}{2}R = \frac{3}{2}R$.Given $n_1 = 2$ moles of oxygen and $n_2 = 4$ moles of argon:$U = n_1 C_{v1} T + n_2 C_{v2} T$$U = 2 	imes (\frac{5}{2}R)T + 4 	imes (\frac{3}{2}R)T$$U = 5RT + 6RT = 11RT$.

Components	$\%$ of the total cellular mass
$A$. Water	$1$. $70-90$
$B$. Protein	$2$. $10-15$
$C$. Carbohydrates	$3$. $3$
$D$. Lipids	$4$. $2$
$E$. Nucleic acids	$5$. $5-7$
$F$. Ions	$6$. $1$

$A$ gas mixture consists of $2$ moles of oxygen and $4$ moles of argon at temperature $T$. Neglecting all vibrational modes,the total internal energy of the system is (in $, RT$)

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