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A vessel contains a mixture of \mathrm{~g} of nitrogen and 11 \mathrm{~g} of carbon dioxide at temperature \mathrm{T}=290 \mathrm{~K}.  If pressure of the mixture \mathrm{P}=1 \mathrm{~atm}, then its density is (\mathrm{R}=8.31 \mathrm{~J} / \mathrm{mol} \mathrm{K}).

Option: 1

1 \mathrm{~kg} / \mathrm{m}^{3}


Option: 2

1.5 \mathrm{~kg} / \mathrm{m}^{3}


Option: 3

2 \mathrm{~kg} / \mathrm{m}^{3}


Option: 4

3 \mathrm{kg} / \mathrm{m}^{3}


Answers (1)

best_answer

As molecular weights of \mathrm{N}_{2} and \mathrm{CO}_{2} are 28 and 44 , and n=(\mathrm{m} / \mathrm{M}),

\mathrm{n}_{\mathrm{N}}=\frac{7}{28}=\frac{1}{4} \text { and } \mathrm{n}_{\mathrm{C}}=\frac{11}{44}=\frac{1}{4}
\text { so } \quad \mathrm{n}=\mathrm{n}_{\mathrm{N}}+\mathrm{n}_{\mathrm{C}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}

Now as according to gas equation \mathrm{PV}=\mathrm{nRT},

\mathrm{\mathrm{V}=\frac{n R T}{P}=\left(\frac{1}{2}\right) \frac{8.31 \times 290}{1.01 \times 10^{5}}=1.19 \times 10^{-2} \mathrm{~m}^{3}}

Mass of the gas is given by

\mathrm{\mathrm{m}=7+11=18 \mathrm{~g}=18 \times 10^{-3} \mathrm{~kg}}

\mathrm{\therefore \quad \rho=(\mathrm{m} / \mathrm{V})=\left(18 \times 10^{-3}\right) /\left(1.19 \times 10^{-2}\right)=1.5 \mathrm{~kg} / \mathrm{m}^{3}}

Posted by

Gautam harsolia

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