Question

If the pressure of 250 cc of dry O2 measured

at 700 mm of Hg & at const temp be raised to

875 mm, what volume will the gas occupy​

Answer 1

Given :

Initial pressure = 700 mm of Hg

Initial volume = 250 cc

Final pressure = 875 mm of Hg

To Find :

We have to find final volume of gas$.$

Concept :

★ Boyle's law : At constant temperature, the pressure of a fixed amount of a gas varies inversely with its volume.

• P ∝ 1/V
• P₁V₁ = P₂V₂

Calculation :

➠ P₁V₁ = P₂V₂

➠ (700)(250) = 875V₂

➠ V₂ = 175000/875

➠ V₂ = 200 cc

Cheers!

Answer 2

Answer :-

To Find :-

The Volume that the gas will apply.

Given :-

• Initial Pressure ($p_{1}$) = 700 mm

• Final Pressure ($p_{2}$) = 875 mm

• Initial Volume ($v_{1}$) = 250 cc

We know :-

⠀⠀⠀⠀⠀⠀⠀⠀⠀Boyle's Law :-

$\boxed{\underline{\bf{P_{1}V_{1} = P_{2}V_{2}}}}$

Where :-

• $P_{1}$ = Initial Pressure

• $P_{2}$ = Final Pressure

• $V_{1}$ = Initial Volume

• $V_{2}$ = Final Volume

Concept :-

From the Boyle's law , we get the relation for Final Volume ($V_{2}$) :

$:\implies \bf{P_{1}V_{1} = P_{2}V_{2}} \\ \\ \\ :\implies \bf{\dfrac{P_{1}V_{1}}{V_{2}} = P_{2}} \\ \\ \\ :\implies \bf{\dfrac{1}{V_{2}} = \dfrac{P_{2}}{P_{1}V_{1}}} \\ \\ \\ :\implies \bf{V_{2} = \dfrac{P_{1}V_{1}}{P_{2}}} \\ \\ \\ \therefore \purple{\bf{V_{2} = \dfrac{P_{1}V_{1}}{P_{2}}}}$

Hence, the Relation for $V_{2}$ is $\bf{V_{2} = \dfrac{P_{1}V_{1}}{P_{2}}}$

Now , we know that 1 cc = 1 ml .

Hence, the Initial Volume is 250 ml.

Solution :-

Using the formula and substituting the values in it , we get :-

$:\implies \bf{V_{2} = \dfrac{P_{1}V_{1}}{P_{2}}} \\ \\ \\ :\implies \bf{V_{2} = \dfrac{700 \times 250}{875}} \\ \\ \\ :\implies \bf{V_{2} = \dfrac{175000}{875}} \\ \\ \\ :\implies \bf{V_{2} = 200} \\ \\ \\ \therefore \purple{\bf{V_{2} = 200 ml}}$

Hence, the Final Volume is 200 ml.