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Do you have enough information to work out the area of the shaded quadrilateral?

Napoleon's Hat

Three equilateral triangles ABC, AYX and XZB are drawn with the point X a moveable point on AB. The points P, Q and R are the centres of the three triangles. What can you say about triangle PQR?

Plane to See

P is the midpoint of an edge of a cube and Q divides another edge in the ratio 1 to 4. Find the ratio of the volumes of the two pieces of the cube cut by a plane through PQ and a vertex.

Diamonds Aren't Forever

Age 16 to 18
Challenge Level

1a) We are initially given a large amount of data which is in non-standard units. Before attempting the main part of the question, it is useful to convert these quanitites into standard units:

$V = 55000\ $cm$^3 = 0.055\ $m$^3$
$P = 900\ $mmHg$\, = 1.18\ $atm$\, = 1.2\ $bar$\, = 1.20 \times 10^5\ $Pa
$T = 49^\circ $C$\, = 322\ $K

Now, we wish to find the number of moles. Using the Ideal Gas law:

$pV = nRT$
$n = \frac{pV}{RT} = \frac{ 1.20 \times 10^5 \times 0.055}{8.314 \times 322} = 2.47\ $moles

b) We are given a density in units of mg/m$^3$, and so it is best to first convert our number of moles into a mass in mg.

Mass = $ 2.47 \times 12.01 = 29.6\ $g$\, = 29600\ $mg

$\therefore V = \frac{29600}{3.52 \times 10^9} = 8.41 \times 10^{-6}\ $m$^3$

Since $V = \frac{4}{3} \pi r^3$
$\rightarrow r = \sqrt[3]{\frac{3V}{4\pi}} = 0.0126\ $m$\, = 1.26\ $cm

2a) $\frac{P_1}{P_2} = \frac{T_1}{T_2}$

Therefore, $P_2 = \frac{P_1T_2}{T_1}$
$= \frac{(1.20 \times 10^5)(253)}{322} = 9.43 \times 10^4$ Nm$^{-2}$

b) Since the volume is doubled and $R$, $T$ and $n$ are constant, the pressure must halve.

$P_3 = \frac{P_2}{2} = 4.71 \times 10^4$ Pa

c) If the gas is not expanding against a vaccuum, then it will do work against the external pressure. Thus, the energy of the gas will decrease, and so its temperature will fall. Since $T$ decreases, this means that $P$ will have to decrease an additional amount. The final pressure is lower than in the previous scenario..

3a) We are back to the original scenario again. The volume and temperature are kept constant, and so we can write:

$\frac{P_1}{n_1} = \frac{P_2}{n_2}$
$P_2 = \frac{n_2P_1}{n_1} = \frac{(2.47 + 5)}{2.47} \times 1.20 \times 10^5 = 3.63 \times 10^5$ Pa

b) We have inserted 5 moles of air into the container. If air is 0.93% argon by volume, then:

$n_{Ar} = 0.0093 \times 5$

The partial pressure of argon, $P_{Ar}$ is given by the mole fraction of argon, $x_{Ar}$, multiplied by the total pressure of the container, $P_T$:

$P_{Ar} = x_{Ar}P_T = \frac{n_{Ar}}{n_T} \times P_T$
$=\frac{0.0093 \times 5}{5+2.47} \times 3.63 \times 10^5 = 2.26 \times 10^3$ Pa

c)We first calculate the concentration of argon molecules by using the ideal gas equation:

$\frac{n}{V} = \frac{P}{RT} = \frac{2260}{8.314 \times 322} = 0.84$ moles/m$^3$

We multiply by Avogadro's number to give the number of molecules per cm$^3$:

$\frac{N}{V} = 0.84 \times 6.022 \times 10^{23} = 5.082 \times 10^{23}$ molecules / m$^3$.

The reciprocal of this concentration gives the volume occupied per molecule!

$\therefore \frac{1}{5.082 \times 10^{23}} = 1.97 \times 10^{-24}$ m$^3$/molecule.

4) This part of the problem is obviously quite open ended, in that a number of different approaches may be taken. We could model such a situation by assuming that the gas as compressible until the particles are all touching. Additionally, we model each particle as a cube:

2.47 moles of carbon atoms contains 14.87 x 10$^{23}$ molecules.

The radius of a carbon atom is ~ 77pm, and so each carbon atom can be modelled as a cube of side 154pm.

Therefore the volume of a carbon atom is $3.63 \times 10^{-30} $m$^3$, and so the total volume occupied by the carbon atoms is $5.43 \times 10^{-6} $m$^3$

5 moles of air is composed of 30.11 x 10$^{23}$ molecules. Air is largely composed of nitrogen, and so as a broad assumption, we can assume that all of these molecules are nitrogen.

The radius of an N is 75pm, but since nitrogen exists as N$_2$, we should model this as a rectangular prism, with two sides of length 150pm, and one of length 300pm.

Thus, the volume of a nitrogen molecule is $6.75 \times 10^{-30}$ m$^3$

Therefore, the overall volume occupied by nitrogen is $2.03 \times 10^{-5}$ m$^3$.

Summing the volumes for carbon and for nitrogen gives the total occupied volume, which is $2.6 \times 10^{-5}\ $m$^3$. Using this model, this is roughly the smallest volume that the gas could be compressed to.