A vessel of capacity 600cm³ contains Hydrogen gas at a pressure of 304 cm of Hg. What will be the pressure of Hydrogen gas, when the vessel is connected to another vessel of the volume of 300cm³ capacity.

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Solution

$P_{1}$ = 304 cm of Hg

$V_{1}$ = 600cm³

now let

$V_{2}$ = 600 + 300= 900cm³

$P_{2}$ = ?

The relationship for Boyle’s Law can be expressed as follows: $P_{1} V_{1} = P_{2} V_{2}$ , where $P_{1}$ and $V_{1}$ are the initial pressure and volume values, and $P_{1}$ and $V_{2}$ are the values of the pressure and volume of the gas after change.

$P_{1} V_{1} = P_{2} V_{2}$

$P_{2} = P_{1} V_{1} / V_{2}$

$P_{2}$ = 600 × 304 / 900

$P_{2}$ = 202.67 cm of Hg
The pressure of Hydrogen gas is 202.67 cm of Hg

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A vessel of capacity 600cm3 contains Hydrogen gas at a pressure of 304 cm of Hg. What will be the pressure of Hydrogen gas, when the vessel is connected to another vessel of the volume of 300cm3 capacity.

A vessel of capacity 600cm³ contains Hydrogen gas at a pressure of 304 cm of Hg. What will be the pressure of Hydrogen gas, when the vessel is connected to another vessel of the volume of 300cm³ capacity.