A gas mixture consists of $3$ moles of oxygen and $5$ moles of argon at temperature $T$ . Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy $(in units of R T)$ of the mixture is :

15

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Solution

The correct option is A $15$
Total energy of the gas mixture,

$E_{m i x} = \frac{f_{1} n_{1} R T_{1}}{2} + \frac{f_{2} n_{2} R T_{2}}{2}$

Where, $f_{1}, f_{2} \to degree of freedom$
$n_{1}, n_{2} \to number of moles$

Here, $f_{1} = 5 (Oxygen-Diatomic), n_{1} = 3$
$f_{1} = 3 (Argon-Monoatomic), n_{2} = 5$

$\Rightarrow E_{m i x} = 3 \times \frac{5}{2} R T + \frac{5}{2} \times 3 R T$

$\Rightarrow E_{m i x} = 15 R T$ .

Hence, option $(A)$ is correct.

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A gas mixture consists of 3 moles of oxygen and 5 moles of argon at temperature T. Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy (in units of RT) of the mixture is :

The correct option is A 15Total energy of the gas mixture, Emix=f1n1RT12+f2n2RT22 Where, f1,f2→degree of freedom n1,n2→number of moles Here, f1=5 (Oxygen-Diatomic),n1=3 f1=3 (Argon-Monoatomic),n2=5 ⇒Emix=3×52RT+52×3RT ⇒Emix=15RT. Hence, option (A) is correct.

A gas mixture consists of $3$ moles of oxygen and $5$ moles of argon at temperature $T$ . Assuming the gases to be ideal and the oxygen bond to be rigid, the total internal energy $(in units of R T)$ of the mixture is :