1. a learner investigates the relationship between the pressure and volume of an enclosed diatomic gas at 25 °c. he records the volume of the gas for different pressures in the table below:

| pressure (kpa) | volume (cm³) | ( \frac{1}{v} ) (cm⁻³) |
|----------------|--------------|-------------------------|
| 40 | 43 | 0.02 |
| 80 | 27 | 0.04 |
| 100 | 22 | (a) |
| 120 | 18 | (b) |

1.1 write down the name of the gas law being investigated. (1)

answer question 4.2 and question 4.3 on the attached answer sheet.

1.2 two ( \frac{1}{v} ) values in the table, (a) and (b), have not been calculated. calculate these values. (1)

1.3 draw a graph of pressure versus ( \frac{1}{v} ) on the attached answer sheet. (4)

1.4 use the graph to determine the volume of the gas at 88 kpa. (2)

1.5 the mass of the enclosed diatomic gas is ( 2.49 imes 10^{-2} ) g.

4.5.1 use the conditions at a pressure of 100 kpa and calculate the molar mass of the enclosed gas. (6)

4.5.2 write down the molecular formula of the enclosed gas. (1)

1.6 the sketch graph below shows the relationship between volume and temperature for an ideal gas.

graph: volume (cm³) on y - axis, temperature (k) on x - axis, straight line through origin

4.6.1 redraw the above graph in the answer book. on the same set of axes, use a broken line to sketch the graph that will be obtained for the diatomic gas above. (1)

4.6.2 fully explain why this diatomic gas deviates from ideal behaviour.

Question

1. a learner investigates the relationship between the pressure and volume of an enclosed diatomic gas at 25 °c. he records the volume of the gas for different pressures in the table below:

| pressure (kpa) | volume (cm³) | ( \frac{1}{v} ) (cm⁻³) |
|----------------|--------------|-------------------------|
| 40             | 43           | 0.02                    |
| 80             | 27           | 0.04                    |
| 100            | 22           | (a)                     |
| 120            | 18           | (b)                     |

1.1 write down the name of the gas law being investigated. (1)

answer question 4.2 and question 4.3 on the attached answer sheet.

1.2 two ( \frac{1}{v} ) values in the table, (a) and (b), have not been calculated. calculate these values. (1)

1.3 draw a graph of pressure versus ( \frac{1}{v} ) on the attached answer sheet. (4)

1.4 use the graph to determine the volume of the gas at 88 kpa. (2)

1.5 the mass of the enclosed diatomic gas is ( 2.49 	imes 10^{-2} ) g.

4.5.1 use the conditions at a pressure of 100 kpa and calculate the molar mass of the enclosed gas. (6)

4.5.2 write down the molecular formula of the enclosed gas. (1)

1.6 the sketch graph below shows the relationship between volume and temperature for an ideal gas.

graph: volume (cm³) on y - axis, temperature (k) on x - axis, straight line through origin

4.6.1 redraw the above graph in the answer book. on the same set of axes, use a broken line to sketch the graph that will be obtained for the diatomic gas above. (1)

4.6.2 fully explain why this diatomic gas deviates from ideal behaviour.

Accepted Answer

### 1.1
# Brief Explanations:
The experiment investigates the relationship between pressure and volume of a gas at constant temperature (25°C). Boyle's Law states that for a fixed amount of gas at constant temperature, the pressure and volume are inversely proportional ($P \propto \frac{1}{V}$ or $PV = \text{constant}$). So the gas law is Boyle's Law.
# Answer:
Boyle's Law

### 1.2
# Explanation:
## Step1: Calculate (a) for $V = 22 \, \text{cm}^3$
To find $\frac{1}{V}$, we take the reciprocal of the volume. So for $V = 22 \, \text{cm}^3$, $\frac{1}{V}=\frac{1}{22}\approx 0.0455 \, \text{cm}^{-3}$ (or more precisely $\frac{1}{22}$).
## Step2: Calculate (b) for $V = 18 \, \text{cm}^3$
For $V = 18 \, \text{cm}^3$, $\frac{1}{V}=\frac{1}{18}\approx 0.0556 \, \text{cm}^{-3}$ (or $\frac{1}{18}$).
# Answer:
(a) $\boldsymbol{\frac{1}{22}\approx 0.0455 \, \text{cm}^{-3}}$ (or $\frac{1}{22}$)
(b) $\boldsymbol{\frac{1}{18}\approx 0.0556 \, \text{cm}^{-3}}$ (or $\frac{1}{18}$)

### 1.3
(Note: Since this is a drawing task, here's a description of how to draw the graph)
# Brief Explanations:
1. **Axes Setup**: Label the x - axis as $\frac{1}{V} \, (\text{cm}^{-3})$ and the y - axis as Pressure (kPa).
2. **Plotting Points**: For each pressure - $\frac{1}{V}$ pair:
   - (40 kPa, $\frac{1}{43}\approx0.0233 \, \text{cm}^{-3}$)
   - (80 kPa, 0.04 $\text{cm}^{-3}$)
   - (100 kPa, $\frac{1}{22}\approx0.0455 \, \text{cm}^{-3}$)
   - (120 kPa, $\frac{1}{18}\approx0.0556 \, \text{cm}^{-3}$)
3. **Drawing the Line**: Connect the plotted points with a straight line (since Boyle's Law implies a linear relationship between $P$ and $\frac{1}{V}$ at constant $T$).
# Answer:
(Graph drawn as per the above instructions. Since it's a drawing, the answer is the graphical representation with the correct axes, labeled points, and a straight line connecting them.)

### 1.4
# Brief Explanations:
1. **Graph Interpretation**: The graph of $P$ vs $\frac{1}{V}$ is a straight line (from Boyle's Law). To find the volume at $P = 88 \, \text{kPa}$, we can use the linear relationship. First, find the equation of the line. Let's take two points, e.g., (40, $\frac{1}{43}$) and (80, 0.04). The slope $m=\frac{80 - 40}{0.04-\frac{1}{43}}=\frac{40}{\frac{0.04\times43 - 1}{43}}=\frac{40\times43}{1.72 - 1}=\frac{1720}{0.72}\approx2388.89$. The equation is $P - 40=m(\frac{1}{V}-\frac{1}{43})$. When $P = 88$, $88 - 40 = 2388.89(\frac{1}{V}-\frac{1}{43})$, $\frac{48}{2388.89}=\frac{1}{V}-\frac{1}{43}$, $\frac{1}{V}=\frac{48}{2388.89}+\frac{1}{43}\approx0.0201 + 0.0233 = 0.0434$, so $V=\frac{1}{0.0434}\approx23 \, \text{cm}^3$ (approximate, can also be done by interpolation on the graph).
2. **Alternative (Graphical)**: On the drawn graph, find the pressure value of 88 kPa on the y - axis, draw a horizontal line to the graph line, then drop a vertical line to the x - axis ($\frac{1}{V}$) and take the reciprocal to get $V$.
# Answer:
Approximately $\boldsymbol{23 \, \text{cm}^3}$ (the value may vary slightly depending on the graph drawing and interpolation method)

### 4.5.1
# Explanation:
## Step1: Recall the Ideal Gas Law
The Ideal Gas Law is $PV = nRT$, where $n=\frac{m}{M}$ ( $m$ is mass, $M$ is molar mass). So $PV=\frac{m}{M}RT$, and we can solve for $M$: $M=\frac{mRT}{PV}$.
## Step2: Identify the values
- $P = 100 \, \text{kPa}=100\times10^{3} \, \text{Pa}$
- $V = 22 \, \text{cm}^3=22\times10^{-6} \, \text{m}^3$
- $m = 2.49\times10^{-2} \, \text{g}=2.49\times10^{-5} \, \text{kg}$
- $R = 8.31 \, \text{J/(mol·K)}$
- $T = 25^{\circ}\text{C}+273 = 298 \, \text{K}$
## Step3: Substitute into the formula
First, calculate $PV$: $PV=100\times10^{3}\times22\times10^{-6}=2.2 \, \text{Pa·m}^3 = 2.2 \, \text{J}$ (since $1 \, \text{Pa·m}^3 = 1 \, \text{J}$)
Then, $mRT=(2.49\times10^{-5} \, \text{kg})\times8.31 \, \text{J/(mol·K)}\times298 \, \text{K}\approx2.49\times10^{-5}\times2476.38\approx0.0617 \, \text{kg·J/(mol·K)·K}=0.0617 \, \text{J·kg/mol}$ (wait, no, units: $m$ in kg, $R$ in J/(mol·K), $T$ in K, so $mRT$ has units $kg\times\frac{J}{mol\cdot K}\times K=\frac{kg\cdot J}{mol}$, and $PV$ has units $J$, so $M=\frac{mRT}{PV}$ has units $\frac{kg\cdot J/mol}{J}=\frac{kg}{mol}=g/mol$ (when we convert $m$ to grams: $m = 2.49\times10^{-2} \, \text{g}$). Let's redo with $m$ in grams:
$m = 2.49\times10^{-2} \, \text{g}$, $R = 0.0821 \, \text{L·atm/(mol·K)}$ or $8.31 \, \text{J/(mol·K)}$, but let's use $R = 8.31 \, \text{J/(mol·K)}$, $V = 22\times10^{-6} \, \text{m}^3$, $P = 100\times10^{3} \, \text{Pa}$, $T = 298 \, \text{K}$
$n=\frac{PV}{RT}=\frac{100\times10^{3}\times22\times10^{-6}}{8.31\times298}=\frac{2.2}{2476.38}\approx8.88\times10^{-4} \, \text{mol}$
Then $M=\frac{m}{n}=\frac{2.49\times10^{-2} \, \text{g}}{8.88\times10^{-4} \, \text{mol}}\approx28 \, \text{g/mol}$
# Answer:
The molar mass of the gas is approximately $\boldsymbol{28 \, \text{g/mol}}$

### 4.5.2
# Brief Explanations:
The molar mass is approximately 28 g/mol, and the gas is diatomic. The molar mass of $N_2$ (nitrogen gas) is $2\times14 = 28 \, \text{g/mol}$, so the molecular formula is $N_2$.
# Answer:
$\boldsymbol{N_2}$

### 4.6.1
(Note: Since this is a drawing task, here's a description)
# Brief Explanations:
1. **Redraw the Original Graph**: Draw the x - axis as Temperature (K) and y - axis as Volume ($\text{cm}^3$). The original graph (for ideal gas) is a straight line passing through the origin (since $V\propto T$ for ideal gas at constant $P$, $V = kT$).
2. **Sketch the Diatomic Gas Graph (Broken Line)**: Real (diatomic) gases deviate from ideal behavior at low temperatures (or high pressures). At low temperatures, the intermolecular forces become significant, and the volume of the gas is larger than that of an ideal gas (because the ideal gas model ignores intermolecular forces). So the graph for the diatomic gas will be above the ideal gas graph at lower temperatures (and will deviate more as temperature decreases). So draw a broken line that starts at the origin (or near it) and curves upwards (or is above the ideal gas line) at lower temperatures.
# Answer:
(Graph redrawn with the original ideal gas line as a solid line and the diatomic gas line as a broken line, with the diatomic gas line deviating above the ideal gas line at lower temperatures.)

### 4.6.2
# Brief Explanations:
1. **Ideal Gas Assumptions**: Ideal gases assume that gas particles have no intermolecular forces and negligible volume compared to the container volume.
2. **Deviation Reasons for Diatomic Gas**:
   - **Intermolecular Forces**: Diatomic gases (like $N_2$) have intermolecular forces (London dispersion forces). At low temperatures (or high pressures), the kinetic energy of the molecules is low, so these intermolecular forces (attractive forces) become significant. This causes the molecules to be pulled closer together than in the ideal gas model, but wait, no - actually, when considering volume vs temperature at constant pressure (Charles's Law - like situation here), at low temperatures, the real gas volume is larger than ideal? Wait, no, let's correct: For real gases, at low temperatures (below the Boyle temperature) and moderate pressures, the intermolecular attractive forces cause the gas to be more compressible (volume less than ideal), but when considering $V$ vs $T$ at constant $P$, the real gas (diatomic) will have a volume that is slightly larger than the ideal gas at lower temperatures because the ideal gas model ignores the volume of the gas molecules themselves. Wait, the two main deviations from ideal gas are:
     - **Volume of Gas Particles**: Real gas particles have a finite volume. In the ideal gas law, $V$ is the volume of the container, but for real gases, the available volume for the gas to move is $V - nb$ (where $n$ is moles, $b$ is a constant related to particle volume). So at low temperatures (when the volume of the container is not much larger than the total volume of the particles), the real gas volume is larger than the ideal gas volume (since the ideal gas assumes $V$ is the free volume, but real gas has less free volume, but wait, no - the ideal gas law $PV = nRT$ assumes $V$ is the volume of the container, and the particles have no volume. So for real gas, the actual free volume is $V - nV_m$ ( $V_m$ is molar volume of particles). So when we plot $V$ (container volume) vs $T$ at constant $P$, the real gas will have a $V$ that is larger than the ideal gas $V$ (because the ideal gas would have $V=\frac{nRT}{P}$, but real gas has $P(V - nV_m)=nRT$, so $V=\frac{nRT}{P}+nV_m$, so $V$ is larger than ideal).
     - **Intermolecular Forces**: At low temperatures, the attractive intermolecular forces (van der Waals forces) between diatomic molecules become significant. These forces cause the molecules to "stick" together slightly, which would tend to decrease the volume, but the effect of the finite particle volume is usually more significant at low temperatures for diatomic gases like $N_2$ (or maybe the other way around - it depends on the gas). However, the main reasons for deviation are:
       - Real gas molecules have a definite volume (they occupy space), so the actual volume available for the gas to expand is less than the container volume, but when plotting $V$ (container volume) vs $T$, the real gas will have a larger $V$ than ideal at the same $T$ and $P$ because the ideal gas law does not account for the volume of the molecules.
       - Real gas molecules experience intermolecular forces (attractive or repulsive). At low temperatures, attractive forces are dominant, which would tend to make the volume smaller, but the finite volume effect can be more significant. For diatomic gases, at temperatures not too low, the finite volume effect and intermolecular forces both cause deviations. In the case of the $V$ vs $T$ graph (constant $P$), the real gas (diatomic) will deviate from the ideal gas (which has a straight line through the origin) because:
         - The ideal gas law assumes no intermolecular forces and no molecular volume. Real diatomic gases have both. At low temperatures, the intermolecular attractive forces cause the gas to condense (or come closer), but if we are still in the gaseous phase, the volume of the real gas will be different from the ideal. The graph for the real gas will not be a perfect straight line through the origin because the intermolecular forces and molecular volume affect the relationship $V\propto T$ (Charles's Law) which holds strictly for ideal gases.
# Answer:
Ideal gases assume gas particles have **no intermolecular forces** and **negligible volume** compared to the container. Diatomic gases deviate because:
1. **Molecular Volume**: Real gas particles (diatomic molecules) have a finite volume. The ideal gas law ignores this, so the actual free volume for gas movement is less than the container volume, altering the $V - T$ relationship.
2. **Intermolecular Forces**: Diatomic gases experience intermolecular forces (e.g., London dispersion forces). At low temperatures (or high pressures), these attractive forces become significant, causing molecules to interact (e.g., "stick" slightly), which violates the ideal gas assumption of independent particles.

These factors mean the real (diatomic) gas does not follow $V\propto T$ (Charles’s Law) strictly, so its $V - T$ graph deviates from the ideal gas’s straight - line (through - origin) behavior.

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QUESTION IMAGE

Question

Explanation:

1.1

Step1: Calculate (a) for \(V = 22 \, \text{cm}^3\)

Step2: Calculate (b) for \(V = 18 \, \text{cm}^3\)

Answer:

1.2