Question

A gas occupies 600 cm cube under a pressure of 700 mm of hg . Find under what pressure the volume of gas will be reduced by 20 percent of its original volume, the temperature remaining constant throughout

Answer 1

Answer:

The volume of the gas will be reduced by 20% of its original volume at 875 mm Hg.

Explanation:

Statement and mathematical equation of Boyle's Law:

Boyle's law states that:
"Under constant temperature conditions, the volume of a fixed mass of gas will be directly proportional to (1/P) of the gas system".

Mathematically, it can be written as:
$V \propto \frac{1}{P}$ [when temperature and mass of the gas are constant]

$\Rightarrow V = \frac{K}{P}$ [K = proportionality constant]

$\Rightarrow PV = K$..................(I)

If we compare two or more gaseous systems, then:

P₁V₁ = P₂V₂ = P₃V₃ =.........= PₙVₙ = K..........(II)

Calculation:

Given that,
The initial volume of the gas (V₁) = 600 cm³
The initial pressure exerted by the gas (P₁) = 700 mm Hg

Now, it is written that the final volume of the gas will be reduced by 20% of its original volume.
∴ The final volume of the gas
$= V_{2} = [ 600 - (\frac{20}{100} \times 600)] = [600 - 120] = 480 cm^{3}$

Let the final pressure of the gas system (P₂) be x.

∴ By the problem,

$P_{1}V_{1} = P_{2}V_{2} \\\Rightarrow 700 \times 600 = x \times 480 \\\Rightarrow x = \frac{700 \times 600}{480} \\\Rightarrow x = 875$

Conclusion:

The final pressure of the gaseous system will be 875 mm Hg.

For more information on Boyle's Law, you may refer to the links below:

State Boyle's Law:
https://brainly.in/question/14698053

State Boyle's law and Charles' law.
https://brainly.in/question/14308092

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