Question

4.1 a small group of learners investigate the relationship between pressure and volume of an enclosed mass of gas at 298 k. they record the volume of the gas for different pressures in the table below:

pressure (kpa)	volume (cm³)	$\frac{1}{v}$ (cm⁻³)
160	27	0.04
200	22	(a)
240	18	(b)

4.1.1 write down the name of the law being investigated.
4.1.2 two $\frac{1}{v}$ values, (a) and (b) have been left out of the table. calculate these values.
4.1.3 draw a graph of pressure against $\frac{1}{v}$ on the graph paper.
4.1.4 use the graph to determine the volume of the gas at 170 kpa.
4.1.5 calculate the slope of the graph.
4.1.6 write down the units of the slope calculated in 4.1.5.

Explanation:

Step1: Calculate value (a)

Given volume $V = 22\ cm^{3}$, then $\frac{1}{V}=\frac{1}{22}\ cm^{-3}\approx0.045\ cm^{-3}$

Step2: Calculate value (b)

Given volume $V = 18\ cm^{3}$, then $\frac{1}{V}=\frac{1}{18}\ cm^{-3}\approx0.056\ cm^{-3}$

Step3: Determine volume at 170 kPa

First, find the equation of the line of the graph of pressure $P$ against $\frac{1}{V}$. Since $P$ is proportional to $\frac{1}{V}$ (Boyle's law $P_1V_1 = P_2V_2$ or $P = k\frac{1}{V}$), we can use two - point form to find the relationship. Let's assume two points $(P_1,\frac{1}{V_1})=(80,0.02)$ and $(P_2,\frac{1}{V_2})=(160,0.04)$. The slope $k=\frac{P_2 - P_1}{\frac{1}{V_2}-\frac{1}{V_1}}=\frac{160 - 80}{0.04 - 0.02}=4000$. The equation of the line is $P = 4000\frac{1}{V}$. When $P = 170\ kPa$, then $170=4000\frac{1}{V}$, and $V=\frac{4000}{170}\ cm^{3}\approx23.5\ cm^{3}$

Step4: Calculate slope of the graph

Using two points $(P_1,\frac{1}{V_1})=(80,0.02)$ and $(P_2,\frac{1}{V_2})=(160,0.04)$
The slope $m=\frac{P_2 - P_1}{\frac{1}{V_2}-\frac{1}{V_1}}=\frac{160 - 80}{0.04 - 0.02}=4000$

Step5: Find units of the slope

The units of pressure is $kPa$ and the units of $\frac{1}{V}$ is $cm^{-3}$. So the units of the slope is $kPa\cdot cm^{3}$

Answer:

4.1.1: Boyle's law
4.1.2: (a) $0.045\ cm^{-3}$, (b) $0.056\ cm^{-3}$
4.1.3: (Graph - not provided in text format, but should be a straight - line graph with pressure on the y - axis and $\frac{1}{V}$ on the x - axis)
4.1.4: $23.5\ cm^{3}$
4.1.5: $4000$
4.1.6: $kPa\cdot cm^{3}$