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A vessel contains $1.60g$ of oxygen and $2.80g$ of nitrogen. The temperature is maintained at $300K$and the volume of the vessel is$0.166{m^3}$. Find the pressure of the mixture.

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Last updated date: 21st Jul 2024
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Answer
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Hint: It's important to remember that the solution is dependent on pressure, which is represented by P, and volume, which is represented by V. We must understand that deductions and conclusions based on observations are states of nature, such as laws of physics, which are based on repeated experiments and or observation.

Complete step-by-step solution:We must understand that temperature plays an important role in the solution. The ideal gas equation is used to determine the answer to the above question.
$PV = nRT$
The ideal gas equation is the one shown above. The ideal gas equation holds valid for all gases in the universe.
Where,
$P = $Pressure
$V = $Volume
$n = $Amount of substance
$R = $Gas constant
$T = $Temperature
Pressure into volume is equal to ideal gas constant into temperature into sum of material, according to the ideal gas equation.
Given:
$V = 0.166{m^3}$
$T = 300K$
Mass of ${O_2} = 1.60g$
${M_{oxygen}} = 32g$
${n_{oxygen}} = \dfrac{{1.60}}{{32}} = 0.05$
Mass of ${N_2} = 2.80g$
${M_{nitrogen}} = 28g$
${n_{nitrogen}} = \dfrac{{2.80}}{{28}} = 0.1$
The partial pressure of oxygen is given by:
${P_{oxygen}} = \dfrac{{{n_{oxygen}}RT}}{V} = \dfrac{{0.05 \times 8.3 \times 300}}{{0.166}} = 750$
The partial pressure of nitrogen is given by:
${P_{nitrogen}} = \dfrac{{{n_{nitrogen}}RT}}{V} = \dfrac{{0.1 \times 8.3 \times 300}}{{0.166}} = 1500$
The sum of the partial pressures is the total pressure.
$ \Rightarrow P = {P_{nitrogen}} + {P_{oxygen}} \\
   \Rightarrow P = 750 + 1500 \\
   \Rightarrow P = 2250Pa \\ $

Note:We should also note that there have been several pressure laws proposed, including Boyle's law, which states that the pressure of an ideal gas is inversely proportional to the volume of the system at constant temperature.