Question

A 20 L container holds 0.650 mol of He gas at 37°C at a pressure of 628.3 bar. What will be new pressure inside the
container if the volume is reduced to 12L The temperature is increased to 177 C and 1.25 mol of additional He gas was added to it?
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Answer 1

Explanation:

Here given,

initial volume, V1 = 20Litres

final volume, V2 = 12 Litres

initial no of moles, n1 = 0.650 mol

final no of moles, n2 = (1.25 + 0.650) mol = 1.9 mol

initial temperature, T1 = 37°C = 310K

final temperature, Tf = 177°C = 450K

initial pressure , P1 = 628.3 bar

final pressure , P2 = ?

Applying formula,

P1V1/n1T1 = P2V2/n2T2

or, (628.3 × 20)/(0.650 × 310) = (P2 × 12)/(1.25 × 450)

or, P2 = (628.3 × 20 × 1.9 × 450)/(0.65 × 310 × 12)

= 4443.3 bar

Hence, final pressure of gas in container will be 4443.3 bar.

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