Question

750 ml of gas when taken in a vessel has a pressure equal to 900 mm of Hg, 1200 ml of O2 gas when taken in another vessel has a pressure equal to 1450 mm of Hg. If both the gases are taken in 1000 ml vessel, what will be the total pressure exerted by the mixture of above gases? Assume that the gases are non-reacting.


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Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of Combined Gas Law has been used. We know that, when temperature is constant then the product of pressure and volume of reactants is equal to the product of pressure and volume of final products. Using this firstly, we can find out seperately the pressure of Nitrogen and Oxygen gas. Then we will combine them to find the total pressure.

Let's do it !! ________________________________________________

★ Formulae Used :-

$\\\;\boxed{\sf{\pink{P_{1}\:V_{1}\;=\;\bf{P_{2}\:V_{2}}}}}$

$\\\;\boxed{\sf{\pink{P_{Total}\;=\;\bf{P_{1}\;+\;P_{2}\;+\; ... \;+\;P_{n}}}}}$

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★ Solution :-

Given,

• For N₂ -

» Initial Pressure = P₁ = 900 mm of Hg

» Initial Volume = V₁ = 750 mL

» Final Volume = V₂ = 1000 mL

• Let the final Pressure of N₂ be P₂

• For O₂ -

» Initial Pressure = P₁ = 1450 mm of Hg

» Initial Volume = V₁ = 1200 mL

» Final Volume = V₂ = 1000 mL

• Let the final Pressure of O₂ be P₂

*Note : The second gas taken is Nitrogen N₂ .

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~ For the Final Pressure of N₂ and O₂ ::

We know that,

$\\\;\sf{\rightarrow\;\;P_{1}\:V_{1}\;=\;\bf{P_{2}\:V_{2}}}$

Now using this formula,

• For N₂ ::

By applying values in the formula, we get

$\\\;\sf{\rightarrow\;\;750\:\times\:900\;=\;\bf{P_{2}\:\times\:1000}}$

$\\\;\sf{\rightarrow\;\;P_{2}\;=\;\bf{\dfrac{750\:\times\:900}{1000}}}$

Now cancelling the zeroes, we get

$\\\;\sf{\rightarrow\;\;P_{2}\;=\;\bf{75\:\times\:9}}$

$\\\;\bf{\rightarrow\;\;P_{2}\;=\;\bf{\blue{675\:\;mm\;of\;Hg}}}$

This is the Final Pressure of N₂

• For O₂ ::

By applying the values in the formula, we get

$\\\;\sf{\rightarrow\;\;1450\:\times\:1200\;=\;\bf{P_{2}\:\times\:1000}}$

$\\\;\sf{\rightarrow\;\;P_{2}\;=\;\bf{\dfrac{1450\:\times\:1200}{1000}}}$

Now cancelling the zeroes, we get

$\\\;\sf{\rightarrow\;\;P_{2}\;=\;\bf{145\:\times\:12}}$

$\\\;\bf{\rightarrow\;\;P_{2}\;=\;\bf{\red{1740\:\;mm\;of\;Hg}}}$

This is the Final Pressure of O₂

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~ For Total Pressure exerted by the mixture ::

We see that, the mixture contains N₂ and O₂.

So if we add the final pressure of these both gases, so we can get final pressure exerted by the mixture.

This is given as,

$\\\;\sf{:\Longrightarrow\;\;P_{Total}\;=\;\bf{P_{1}\;+\;P_{2}\;+\; ... \;+\;P_{n}}}$

• Here P₁ shows the final pressure of N₂

• Here P₂ shows the final pressure of O₂

$\\\;\sf{:\Longrightarrow\;\;P_{Total}\;=\;\bf{\blue{P_{1}}\;+\;\red{P_{2}}}}$

By applying values we get,

$\\\;\sf{:\Longrightarrow\;\;P_{Total}\;=\;\bf{\blue{675}\;+\;\red{1740}}}$

$\\\;\bf{:\Longrightarrow\;\;P_{Total}\;=\;\bf{\orange{2415\:\;mm\;of\;Hg}}}$

This is the required answer.

$\\\;\underline{\boxed{\tt{Total\;\:pressure\;\:exerted\;\:by\;\:mixture\;=\;\bf{\purple{2415\;\:mm\;of\;Hg}}}}}$

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★ More to know :-

$\\\;\sf{\leadsto\;\;\dfrac{P_{1}\:V_{1}}{T_{1}}\;=\;\dfrac{P_{2}\:V_{2}}{T_{2}}}$

$\\\;\sf{\leadsto\;\;P\:V\;=\;n\:R\:T}$

$\\\;\sf{\leadsto\;\;V_{t}\;=\;V_{0}\;+\;\dfrac{t}{273.15}\:V_{0}}$

$\\\;\sf{\leadsto\;\;\dfrac{P_{1}}{P_{2}}\;=\;\dfrac{V_{2}}{V_{1}}}$

$\\\;\sf{\leadsto\;\;M\;=\;\dfrac{d\:R\:T}{P}}$