The ideal gas equation is a simple equation which can be used to model certain gases under certain conditions. In this question, assume that it is always applicable. The equation is as follows:

$$pV = nRT$$

where

$p$ = pressure

$V$ = volume

$n$ = number of moles

$R$ is 8.314 JK$^{-1}$mol$^{-1}$

$T$ = temperature

I vaporise a diamond using a laser, such that the gas fills 55000 cm$^3$ at a pressure of 900 mmHg and a temperature of 49$^\circ$C.

How many moles of carbon are there in my original sample of diamond? Note that data for this question are at the bottom of the page.

Assuming that the diamond can be modelled as a sphere of density 3.52 $\times$ 10$^{9}$ mg/m$^3$, what would the radius of my diamond have been in cm?

I now cool the vapour to -20$^\circ$C at a constant volume. There is no other heat transfer to or from the gas.

What is the new pressure in Nm$^{-2}$?

If I allow the volume of the gas to double (against a vacuum), what would the new pressure be?

If the expansion had not been against a vacuum, would the new pressure be larger or smaller than that calculated previously? Why?

The gas is now returned to its original temperature and volume, and 5 moles of air are introduced into the container.

What is the new pressure of the gas in container?

Given that air is 0.93% argon (by volume), what is the partial pressure of argon?

If the argon molecules are evenly distributed in the container, what volume does each molecule occupy?

I now try to compress the gas into the smallest volume possible.

Making suitable modelling assumptions, estimate what this smallest volume would be.

Data:

1 atm = 760 mmHg

1 bar = 10$^5$ Pa = 0.987 atm

N radius = 75 pm

C radius = 77 pm