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A student investigates four gases - OCR Gateway - GCSE Physics - Question 22 - 2019 - Paper 3
Question 22
A student investigates four gases. Look at her data. | Gas | Pressure (Pa) | Volume (m³) | |-----|---------------|--------------| | A | 5 | 0.5 ... show full transcript
Worked Solution & Example Answer:A student investigates four gases - OCR Gateway - GCSE Physics - Question 22 - 2019 - Paper 3
Step 1
Which two readings are for the same mass of the same gas at a constant temperature?
Answer
To determine which two readings are for the same mass of the same gas at a constant temperature, we can apply the ideal gas law relationship, which states that for a constant mass and temperature:
Now we can analyze the provided data. Calculating for each pair:
- For Gas A and Gas B:
eq 10 imes 0.4$$
- For Gas B and Gas C:
eq 20 imes 0.2$$
- For Gas C and Gas D: Therefore, the two readings that are for the same mass of the same gas are Gas C and Gas D.
Step 2
Explain, using ideas about particles, how temperature affects gas pressure.
Answer
Gas pressure is affected by temperature due to the kinetic theory of gases. As temperature increases, the average kinetic energy of the gas particles also increases. This causes the particles to move faster and collide more frequently with the walls of the container.
More collisions contribute to an increase in pressure since pressure is defined as the force exerted per unit area. Therefore, at a constant volume, as temperature rises, the pressure of the gas increases because the higher kinetic energy translates to more forceful and frequent particle impacts on the walls of the container.
Step 3
Calculate the pressure at the bottom of a 0.5m tall measuring cylinder filled with a liquid.
Answer
To calculate the pressure at the bottom of the measuring cylinder, we can use the hydrostatic pressure formula:
ho imes g$$ Where: - $P$ is the pressure at the bottom, - $h$ is the height of the liquid column (0.5 m), - $ ho$ is the density of the liquid (1100 kg/m³), - $g$ is the acceleration due to gravity (9.81 m/s²). Substituting the values: $$P = 0.5 imes 1100 imes 9.81$$ Calculating this gives: $$P = 5395.5 ext{ Pa}$$ Thus, the pressure at the bottom of the cylinder is approximately **5395.5 Pa**.