Question

A gas at pressure 712 mm of Hg has volume 650 cm^3. What will be its volume at 0.8
atm?​

Answer 1

Answer:

Denoting Pressure 1 and Pressure 2 as P1 and P2; Volume 1 and Volume 2 as V1 and V2 respectively.

P1 = 712 mm of Hg

V1 = 650 cc

P2 = 0.8 atm

= 1 atm = 760 mm of Hg

= 0.8 atm = (760*0.8) mm of Hg

= 608 mm of Hg

According to Boyle's law,

P1V1 = P2V2

712*650 = 608V2

(712*650) / 608 = V2

= The New Volume will be 761.18 cc.

I hope this helps.

Answer 2

Given:-

→ Initial pressure = 712 mm Hg

→ Initial volume of the gas = 650 cm³

→ Final pressure = 0.8 atm

To find:-

→ Final volume of the gas.

Solution:-

Firstly, let's convert the initial pressure from mm Hg to atm.

⇒ 1 mm Hg = 1/760 atm

⇒ 712 mm Hg = 712×1/760

⇒ 0.94 atm

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Now according to Boyle's Law, we know that :-

P₁V₁ = P₂V₂ [T is constant]

Where :-

• P₁ is the initial pressure.

• V₁ is the initial volume of the gas.

• P₂ is the final pressure.

• V₂ is the final volume.

Substituting values, we get :-

⇒ 0.94(650) = 0.8(V₂)

⇒ 611 = 0.8V₂

⇒ V₂ = 611/0.8

⇒ V₂ = 763.75 cm³

Thus, required volume of the gas is 763.75cm³ .

Some Extra Information:-

Boyle's Law describes pressure-volume relationship of gases at a constant temperature. It was given by Anglo-Irish scientist Robert Boyle (1662). The law states that :

'The volume of a fixed mass of gas is inveresely proportional to it's pressure at a constant temperature'.