QUESTION 4
QUESTION 4.1
Question 4.1.1 [2 marks]
Write a suitable hypothesis for this investigation.
Hypothesis: As the temperature of the gas increases, the volume of the gas will also increase, demonstrating that there is a direct relationship between temperature and volume.
Question 4.1.2 [2 marks]
Identify the dependent and independent variables.
Independent Variable: Temperature of the gas.
Dependent Variable: Volume of the gas (measured by the diameter of the balloon).
Question 4.1.3 [2 marks]
Identify 2 constant variables.
- The amount of gas (the fixed volume of gas in the balloon).
- The external pressure exerted on the gas (since the balloon is always in the same atmospheric conditions).
QUESTION 4.2
Question 4.2.1 [1 mark]
Which law is being investigated by the grade 11 class?
Answer: The Charles' Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) at constant pressure.
Question 4.2.2 [2 marks]
Give TWO reasons why the graph does not start at the origin.
- A gas cannot occupy zero volume at absolute zero temperature; thus, even if the temperature is zero, the volume cannot be zero.
- The behavior of real gases deviates from ideal gas behavior at low temperatures and high pressures, preventing the graph from passing through the origin.
Question 4.2.3 [3 marks]
Using the values highlighted in red on the graph, find the volume of the balloon (in dm³) when the temperature is 450 K.
(Assuming the graph shows a direct relationship, for example, let's say at 450 K the volume is clearly indicated as X dm³. Since we don't have the specific values visible, the answer should reflect that.)
Answer: The volume of the balloon at 450 K is X dm³. (Replace X with the actual value taken from the graph if it is provided.)
Question 4.2.4 [2 marks]
Explain one difference between a real gas and an ideal gas.
Answer: A real gas does not behave perfectly according to the gas laws due to intermolecular forces and finite molecular volume, particularly under high pressure or low temperature conditions. In contrast, an ideal gas is a hypothetical gas that perfectly follows the gas laws with no intermolecular forces acting upon it, regardless of the conditions.
QUESTION 5
Question 5.1 [5 marks]
Use the values given in the above paragraph to calculate the volume (dm³) of the balloon on the surface of the Earth.
Given Data:
- Mass of He = 141.5 kg
- Molar mass of Helium (He) = 4 g/mol = 0.004 kg/mol
- Standard Pressure (P) = 101.3 kPa
- Standard Temperature (T) = 0°C = 273 K
- Ideal Gas Constant (R) = 8.31 J/(mol·K)
Calculating Moles of Helium: \[ \text{Moles of He} = \frac{\text{mass}}{\text{molar mass}} = \frac{141.5 \text{ kg}}{0.004 \text{ kg/mol}} = 35,375 \text{ mol} \]
Calculating Volume using Ideal Gas Law: \[ PV = nRT \implies V = \frac{nRT}{P} \] Using values: \[ V = \frac{35,375 \text{ mol} \times 8.31 \text{ J/(mol·K)} \times 273 \text{ K}}{101.3 \text{ kPa}} \] (Since 1 kPa = 1000 Pa, convert pressure) \[ = \frac{35,375 \times 8.31 \times 273}{101,300} \] Calculating gives: \[ = \frac{838,814,441.25}{101,300} \approx 8284.01 \text{ m}^3 \] Convert to \(dm^3\) (1 m³ = 1000 dm³): \[ = 8284.01 \times 1000 = 8,284,010 \text{ dm}^3 \]
Answer: The volume of the balloon on the surface of the Earth is approximately 8,284,010 dm³.
Question 5.2 [5 marks]
If it is given that the pressure the balloon experiences at 39 km is 89.5 kPa and the temperature at this height is given as -37℃, calculate the volume of the balloon at this height.
Convert Temperature to Kelvin: \[ T = -37 + 273 = 236 \text{ K} \]
Using the Ideal Gas Law: \[ V = \frac{nRT}{P} \] Substitute the known values: \[ =P \text{ now is in kPa} = 89.5 \text{ kPa} = 89,500 \text{ Pa} \]
Calculating Volume: \[ V = \frac{35,375 \text{ mol} \times 8.31 \text{ J/(mol·K)} \times 236 \text{ K}}{89,500 \text{ Pa}} \] Calculating gives: \[ = \frac{686925219.375}{89,500} \approx 7675.93 \text{ m}^3 \] Convert to \(dm^3\): \[ = 7675.93 \times 1000 = 7,675,930 \text{ dm}^3 \] In scientific notation: \[ = 7.68 \times 10^6 \text{ dm}^3 \]
Answer: The volume of the balloon at 39 km is approximately 7.68 × 10^6 dm³.
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