Date | May 2017 | Marks available | 2 | Reference code | 17M.3.SL.TZ1.1 |
Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
Command term | Suggest | Question number | 1 | Adapted from | N/A |
Question
The equipment shown in the diagram was used by a student to investigate the variation with volume, of the pressure p of air, at constant temperature. The air was trapped in a tube of constant cross-sectional area above a column of oil.
The pump forces oil to move up the tube decreasing the volume of the trapped air.
The student measured the height H of the air column and the corresponding air pressure p. After each reduction in the volume the student waited for some time before measuring the pressure. Outline why this was necessary.
The following graph of p versus \(\frac{1}{H}\) was obtained. Error bars were negligibly small.
The equation of the line of best fit is \(p = a + \frac{b}{H}\).
Determine the value of b including an appropriate unit.
Outline how the results of this experiment are consistent with the ideal gas law at constant temperature.
The cross-sectional area of the tube is 1.3 × 10–3\(\,\)m2 and the temperature of air is 300 K. Estimate the number of moles of air in the tube.
The equation in (b) may be used to predict the pressure of the air at extremely large values of \(\frac{1}{H}\). Suggest why this will be an unreliable estimate of the pressure.
Markscheme
in order to keep the temperature constant
in order to allow the system to reach thermal equilibrium with the surroundings/OWTTE
Accept answers in terms of pressure or volume changes only if clearly related to reaching thermal equilibrium with the surroundings.
[1 mark]
recognizes b as gradient
calculates b in range 4.7 × 104 to 5.3 × 104
Pa\(\,\)m
Award [2 max] if POT error in b.
Allow any correct SI unit, eg kg\(\,\)s–2.
[3 marks]
\(V \propto H\) thus ideal gas law gives \(p \propto \frac{1}{H}\)
so graph should be «a straight line through origin,» as observed
[2 marks]
\(n = \frac{{bA}}{{RT}}\) OR correct substitution of one point from the graph
\(n = \frac{{5 \times {{10}^4} \times 1.3 \times {{10}^{ - 3}}}}{{8.31 \times 300}} = 0.026 \approx 0.03\)
Answer must be to 1 or 2 SF.
Allow ECF from (b).
[2 marks]
very large \(\frac{1}{H}\) means very small volumes / very high pressures
at very small volumes the ideal gas does not apply
OR
at very small volumes some of the assumptions of the kinetic theory of gases do not hold
[2 marks]