Question

A sample of neon occupies a volume of 478 mL at STP. What will be the volume of the neon when the pressure is reduced to 93.3 kPa?

Answer 1

The volume of neon when the pressure is reduced to 93.3 kPa is 519 mL.

Explanation:

The kinetic theory of gases is mostly based on Boyle's law. From the Boyle's law, the pressure experienced by any gas molecules is inversely proportional to volume of the gas molecules. Also this inverse relation is obeyed if and only if the number of moles and temperature of the gas molecules remained constant.

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

So if there is a change in pressure then there will be inverse change in volume. That means if there is decrease in the pressure of gas molecules then there will be increase in the volume and vice versa.

So the Boyle's law is combined as $P_{1} V_{1} = P_{2} V_{2}$

As here the initial pressure or $P_{1}$ is 1 atm or 101.3 kPa and the initial volume is 478 mL. Similarly, the final pressure is 93.3 kPa and the final volume will be

$101.3*10^{3}*478*10^{-3} = 93.3*10^{3} * V_{2}$

$V_{2} = 519 mL$

So, the volume of neon when the pressure is reduced to 93.3 kPa is 519 mL.