Answer
Assuming ideal gas assumption to hold valid, then we know,
$$
\begin{aligned} p \cdot V &=n \cdot R \cdot T \\ p \cdot V &=\left(\frac{m}{M}\right) \cdot R \cdot T \\ M &=\left(\frac{m}{V}\right) \cdot\left(\frac{1}{p}\right) \cdot R \cdot T \\ M &=\frac{\rho \cdot R \cdot T}{p} \end{aligned}
$$
where,
$\mathrm{m}=$ Mass of the gas,
$\mathrm{M}=$ Molar mass of the gas,
$\rho=$ mass density of the gas
From the question we know,
$\rho=1.146 \frac{\mathrm{Kg}}{\mathrm{m}^{3}}=1.146 \frac{\mathrm{g}}{\text { litre }}$
$\mathrm{p}=0.987 \mathrm{bar}=0.974 \mathrm{atm}$
$\mathrm{T}=27^{\circ} \mathrm{C}=300.15 \mathrm{K}$
Substituting the values in the equation,
$$
\begin{aligned} M &=\frac{1.146 \times 0.0821 \times 300.15}{0.974} \mathrm{g} \\ &=29 \mathrm{g} \end{aligned}
$$
Hence the molar mass of the air is 29 $\mathrm{g}$
Let the mole fraction of Oxygen be $x$ and mole fraction of Nitrogen be $1-x$
Assuming only two gases,
$\begin{aligned} \text { Molar mass of gas } &=x \times \text { Molar mass of oxygen }+(1-x) \times \text { Molar mass of Nitrogen } \\ 29 \mathrm{g} &=x \times 32+(1-x) \times 28 \mathrm{g} \\ x &=0.25 \end{aligned}$
Hence the mole fraction of Oxygen is 0.25 and mole fraction of Nitrogen is $0.75 .$
Calculating partial pressure,
$$
\begin{aligned} \text { Partial pressure of a gas } &=\text { Mole fraction of the gas } \times \text { Pressure } \\ \text { Partial pressure of Oxygen } &=0.25 \times 0.987 \text { bar }=0.246 \text { bar } \\ \text { Partial pressure of Nitrogen } &=0.75 \times 0.987 \text { bar }=0.740 \text { bar } \end{aligned}
$$
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The pressure of the system is 0.028 atm and volume is 31 liters
Work Step by Step
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