Question

3. At a pressure of 405 kPa, the volume of a gas is 6.00 cm

Answer 1

The pressure gets increased to 486 kPa from 405 kPa, when the volume is decreased from 6 cm³ to 4 cm³.

Explanation:

In the present problem, the temperature is said to remain at constant and there is change in the pressure. So according to Boyle's law, the relationship between pressure and volume of any gaseous objects are inversely related to each other. In other words, the pressure attained by gas molecules in a container will be inversely proportional to the volume of the gas molecules occupied in the container, at constant temperature.

$V=\frac{1}{P}$

So, if two volumes V₁ and V₂ are considered, then their respective pressure will be represented as P₁ and P₂. Then, as per Boyle's law,

$V_{1}P_{1}=V_{2}P_{2}$

So let us consider, V₁ = 6 cm³ and V₂ = 4 cm³ and pressure P₁ = 405 kPa and we have to determine P₂.

Then,  $6*405=5*P_{2}\\ \\P_{2}=\frac{2430}{5} =486 kPa$

So, the pressure at new volume of 4 cm³ is 486 kPa. It can be seen that as there is decrease in the volume, there is an increase in the pressure. So it satisfied the Boyle's law.

Thus, the pressure gets increased to 486 kPa from 405 kPa, when the volume is decreased from 6 cm³ to 4 cm³.