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  • The pressure exerted by a non-reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27o C is _________kPa. (Round off to the Nearest Integer). [Assume gases are ideal, <img alt="R = 8.314 \; J\; mol^

The pressure exerted by a non-reactive gaseous mixture of 6.4 g of methane and 8.8 g of carbon dioxide in a 10 L vessel at 27o C is _________kPa. (Round off to the Nearest Integer). [Assume gases are ideal, R = 8.314 \; J\; mol^{-1}K^{-1} Atomic masses : C :12.0\; u, H:1.0\; u,O:16.0 \; u]
Option: 1 150
Option: 2 -
Option: 3 -
Option: 4 -

Answers (1)

best_answer

Given,

V = 10 L, T = 27o C = 300 K

and

\\\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\\\textup{or}\\\mathrm{R}=8.314\ \mathrm{m}^{3} \ \mathrm{Pa} \ \mathrm{K}^{-1} \mathrm{mol}^{-1}

So, Convert Volume in cubic meter unit Because R in Pa unit has meter unit.

\mathrm{V = 10\ L = 10\times10^{-3}\ m^3}

And

\mathrm{m}_{\text {methane }}=6.4 \mathrm{~g},\mathrm{~m}_{\mathrm{CO}_{2}}=8.8 \mathrm{~g}

Asking about The pressure exerted by a non-reactive gaseous mixture -

So, we know the formula -

\mathrm{PV}=\mathrm{n}_{\text {total }} \mathrm{RT}

Here, the mixture of gases is also a gas and moles will be total moles.

\text { Total moles of gases, } \mathrm{n}=\mathrm{n}_{\mathrm{CH}_{4}}+\mathrm{n}_{\mathrm{CO}_{2}}

\mathrm{n_{Total}}=\frac{6.4}{16}+\frac{8.8}{44}=0.6

Now,

\mathrm{P}=\frac{\mathrm{n}_{\text {total }} \mathrm{RT}}{\textup{V}}

\textup{P}=\frac{0.6 \times 8.314 \times 300}{10 \times 10^{-3}}

\textup{P}=1.49652 \times 10^{5} \mathrm{~Pa}

\textup{P}=149.652 \mathrm{kPa}

\textup{P}\approx 150 \ \mathrm{kPa}

Ans = 150

Posted by

Kuldeep Maurya

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