Question

if the volume of a given mass at STP is 52m3 . what will be its volume at the pressure 104cm Hg and temperature remaining constant​

Answer 1

Answer:

$The \: final \: volume \: is \: = > 38 \: m {}^{3}$

Explanation:

$Initial \: volume \: \: V _{1} = 52 \: m {}^{3} \\ Initial \: pressure \: \: P _{1} = 76 \: cm \: of \: Hg \\ Final \: pressure \: \: P _{2} = 104 \: cm \: of \: Hg \\ Final \: volume \: = \: ? \\ According \: to \: the \: Boyle's \: law, \\ {P _{1} \: V _{1}} = {P _{2} \: V _{2} } \\ Therefore \: \: V _{2} = \frac{76 \times 52}{104} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = > 38 \: m {}^{3}$

Answer 2

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Given :-

☆volume of given mass of a gas at STP is 52 cubic metre what will be its volume at the pressure of 104 CM of HG temperature remaining constant

Solution :-

P₁V₁ = P₂V₂

(100)(52) = (138.32)(V₂)

5200 = 138.32V₂

5200/138.32 = V₂

37.5 = V₂

The volume of the gas decreases is 37.5 m³.