Question #e7184

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Answer 1 · 2017-07-25T15:17:47

On average, the pressure dropped by 0.83 bar each day.

From the Ideal Gas Law, you know that

$p V = n R T$ $pV = nRT$

Thus,

$p \propto n$ $p ∝ n$ .

The number of moles, $n$ $n$ , is directly proportional to the mass $m$ $m$ of the gas, so

$p \propto m$ $p ∝ m$ .

This means that, if the mass decreases by a certain factor, the pressure will decrease by the same factor.

Here, the mass has decreased to 10 kg from 15 kg, a factor of $\frac{10}{15}$ $10/15$ .

The pressure will decrease by the same factor.

Hence, the new pressure will be

$10 bar \times \frac{10}{15} = 6.67 bar$ $"10 bar" × 10/15 = "6.67 bar"$

The drop in pressure is

$10 bar - 6.67 bar = 3.33 bar$ $"10 bar - 6.67 bar = 3.33 bar"$

This pressure drop occurred over four days.

∴ The average pressure drop per day was

$\frac{3.33 bar}{4 days} = 0.83 bar/day$ $"3.33 bar"/"4 days" = "0.83 bar/day"$