# Question e7184

Jul 25, 2017

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

#### Explanation:

From the Ideal Gas Law, you know that

$p V = n R T$

Thus,

p ∝ n.

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

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 $\frac{10}{15}$.

The pressure will decrease by the same factor.

Hence, the new pressure will be

$\text{10 bar" × 10/15 = "6.67 bar}$

The drop in pressure is

$\text{10 bar - 6.67 bar = 3.33 bar}$

This pressure drop occurred over four days.

∴ The average pressure drop per day was

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