Question

A container of hydrogen at 172 kPa was decreased to 85.0 kPa producing a new volume of 3L. What was the original volume?

Answer 1

Answer:

The answer is 1.48 L

Explanation:

In order to find the original volume we use the same for Boyle's law which is

$P_1V_1 = P_2V_2$

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the original volume

$V_1 = \frac{P_2V_2}{P_1}$

From the question

P1 = 172 kPa = 172000 Pa

P2 = 85 kPa = 85000 Pa

V2 = 3 L

We have

$V_1 = \frac{85000 \times 3}{172000} = \frac{255000}{172000} = \frac{255}{172} \\ = 1.482558139...$

We have the final answer as

1.48 L

Hope this helps you