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  • A vessel contains a mixture of of nitrogen and of carbon di

A vessel contains a mixture of 7 \mathrm{g} of nitrogen and 11 \mathrm{g} of carbon dioxide at temperature T=290 \mathrm{~K}. If pressure of the mixture P=1 atm, calculate its density \mathrm{(R=8.31 \mathrm{~J} / \mathrm{mol}\, \mathrm{k})}

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

molecular weight of \mathrm{N}_{2} and \mathrm{CO}_{2} are 28 and 44 , and \mathrm{n}=\frac{\mathrm{m}}{\mathrm{M}},

\mathrm{\mathrm{n}_{\mathrm{N}}=\frac{7}{28}=\frac{1}{4} \, and \, \mathrm{n}_{\mathrm{C}}=\frac{11}{44}=1 / 4}

\mathrm{So, \mathrm{n}=\mathrm{n}_{\mathrm{N}}+\mathrm{n}_{\mathrm{C}}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} }
Now, according to gas law \mathrm{\mathrm{PV}=\mathrm{nRT} }

\mathrm{\mathrm{V}=\frac{\mathrm{nRT}}{\mathrm{P}}=\left(\frac{1}{2}\right) \frac{8.31 \times 290}{1.01 \times 10^{5}}=1.19 \times 10^{-2} \mathrm{~m}^{3} }
\mathrm{and \, \mathrm{m}=7+11=18 \mathrm{gm}=18 \times 10^{-3} \mathrm{~kg} }

\mathrm{so, p=\frac{m}{V}=\frac{18 \times 10^{-3} \mathrm{~kg}}{1.19 \times 10^{-2} \mathrm{~m}^{3}}=1.5 \mathrm{~kg} / \mathrm{m}^{3} }

Posted by

Sanket Gandhi

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