A mixture of ideal gases has the following composition by mass:

$N_{2}$ $O_{2}$ $C O_{2}$

$60$ % $30$ % $10$ %

If the universal gas constant $8314 J / k m o l - K$ , the characteristic gas constant of the mix
274.86

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Solution

The correct option is A 274.86
$Equation state for N_{2}$ ,

$p_{N_{2}} V = m_{N_{2}} R_{N_{2}} T$

or $\frac{p_{N_{2}} V}{T} = m_{N_{2}} R_{N_{2}}$ \)

Where $R_{N_{2}} = \frac{¯ ¯¯ ¯ R}{M_{N_{2}}}$

$∴ \frac{p_{N_{2}} V}{T} = \frac{m_{N_{2}} ¯ ¯¯ ¯ R}{M_{N_{2}}} . . . (i)$

SImilarly for $O_{2}$ and $C O_{2}$

$\frac{p_{C O_{2}} V}{T} = \frac{m_{O_{2}} ¯ ¯¯ ¯ R}{M_{O_{2}}} . . . . (i i)$

and $\frac{p_{C O_{2}} V}{T} = \frac{m_{C O_{2}} ¯ ¯¯ ¯ R}{M_{C O_{2}}} . . . (i i i)$

$Adding equation (i), (ii) and (iii), we get$

$\frac{V}{T} (p_{N_{2}} + p_{O_{2}} + p_{C O_{2}} +) = \frac{m_{N_{2}} ¯ ¯¯ ¯ R}{M_{N_{2}}} + \frac{m_{O_{2}} ¯ ¯¯ ¯ R}{M_{O_{2}}} + \frac{m_{C O_{2}} ¯ ¯¯ ¯ R}{M_{C O_{2}}}$

$\frac{V}{T} + p = (\frac{m_{N_{2}}}{M_{N_{2}}} + \frac{m_{O_{2}}}{M_{O_{2}}} + \frac{m_{C O_{2}}}{M_{C O_{2}}}) ¯ ¯¯ ¯ R . . . (i v)$

$where, pressure of mixture$

$p = p_{N_{2}} + p_{O_{2}} + p_{C O_{2}}$

$Equation of state for mixture of ideal gas$

$p V = m R T or$

$\frac{p V}{T} = m R . . . (v)$

$Equating ECs (iv) and (v) we get$

$m R = (\frac{m_{N_{2}}}{M_{N_{2}}} + \frac{m_{O_{2}}}{M_{O_{2}}} + \frac{m_{C O_{2}}}{M_{C O_{2}}}) ¯ ¯¯ ¯ R$

$m R = (\frac{0.60 m}{M_{N_{2}}} + \frac{0.30 m}{M_{O_{2}}} + \frac{0.1 m}{M_{C O_{2}}}) ¯ ¯¯ ¯ R$

$or R = (\frac{0.60}{M_{N_{2}}} + \frac{0.30}{M_{O_{2}}} + \frac{0.1}{M_{C O_{2}}}) ¯ ¯¯ ¯ R$

$== (\frac{0.60}{28} + \frac{0.30}{32} + \frac{0.10}{44}) \times 8314$

$= (0.02142 + 0.00937 + 0.00227) \times 8314$

$= 0.03306 \times 8134$

$= 274.86 J / k g K$

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