Question
8 gm of oxygen, 14 gm of nitrogen and 22 gm of carbon dioxide are mixed in a vessel of 10 liters volume and $27^{\circ} C$. Find the pressure exerted by the mixture.
✓
Answer
According to Dalton's law of partial pressure :
$\begin{array}{l}P=P_1+P_2+P_3 \\\because \quad P=\frac{n RT}{V} \text { and } n=\frac{m}{M} \text {, so } P=\frac{m}{M} \frac{RT}{V} \\\therefore \quad P=\frac{RT}{V}\left[\frac{m_1}{M_1}+\frac{m_2}{M_2}+\frac{m_3}{M_3}\right] \\\text { Putting the value }=\frac{8.3 \times 300}{10^{-2}}\left[\frac{8}{32}+\frac{14}{28}+\frac{22}{44}\right] \\\because \quad R=8.3 \text { joule/gm/mole Kelvin } \\
T=273+27=300 K \\V=10 \text { liter }\end{array}$
$\begin{array}{l}=10 \times 1000 \text {cubic} ~ \text{cm} \\=10^4 \text { cubic } cm \\=\frac{10^4}{10^6} \text { cubic } m=10^{-2} \text { cubic } m\end{array}$
and molecular weight of $O _2, N_2$ and $CO _2$ will be 32, 28, 44.
$\begin{aligned}\therefore \quad P & =\frac{8.3 \times 300}{10^{-2}}\left[\frac{1}{4}+\frac{1}{2}+\frac{1}{2}\right] \\& =\frac{8.3 \times 300}{10^{-2}}\left[\frac{1+2+2}{4}\right] \\& =\frac{8.3 \times 300 \times 5}{4 \times 10^{-2}} \\& =3.1 \times 10^5 Newton / m^2 \\& =\frac{3.1 \times 10^5}{1.01 \times 10^5}=3 \text { Atmosphere pressure }\end{aligned}$
$\begin{array}{l}P=P_1+P_2+P_3 \\\because \quad P=\frac{n RT}{V} \text { and } n=\frac{m}{M} \text {, so } P=\frac{m}{M} \frac{RT}{V} \\\therefore \quad P=\frac{RT}{V}\left[\frac{m_1}{M_1}+\frac{m_2}{M_2}+\frac{m_3}{M_3}\right] \\\text { Putting the value }=\frac{8.3 \times 300}{10^{-2}}\left[\frac{8}{32}+\frac{14}{28}+\frac{22}{44}\right] \\\because \quad R=8.3 \text { joule/gm/mole Kelvin } \\
T=273+27=300 K \\V=10 \text { liter }\end{array}$
$\begin{array}{l}=10 \times 1000 \text {cubic} ~ \text{cm} \\=10^4 \text { cubic } cm \\=\frac{10^4}{10^6} \text { cubic } m=10^{-2} \text { cubic } m\end{array}$
and molecular weight of $O _2, N_2$ and $CO _2$ will be 32, 28, 44.
$\begin{aligned}\therefore \quad P & =\frac{8.3 \times 300}{10^{-2}}\left[\frac{1}{4}+\frac{1}{2}+\frac{1}{2}\right] \\& =\frac{8.3 \times 300}{10^{-2}}\left[\frac{1+2+2}{4}\right] \\& =\frac{8.3 \times 300 \times 5}{4 \times 10^{-2}} \\& =3.1 \times 10^5 Newton / m^2 \\& =\frac{3.1 \times 10^5}{1.01 \times 10^5}=3 \text { Atmosphere pressure }\end{aligned}$
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