An apparatus is used to verify a gas law. The glass jar contains a fixed volume of air. Measurements can be taken using the thermometer and the pressure gauge.

The apparatus is cooled in a freezer and then placed in a water bath so that the temperature of the gas increases slowly. The pressure and temperature of the gas are recorded.

a. The graph shows the data recorded.

Identify the fundamental SI unit for the gradient of the pressure–temperature graph.[1]

The experiment is repeated using a different gas in the glass jar. The pressure for both experiments is low and both gases can be considered to be ideal.

(i) Using the axes provided in (a), draw the expected graph for this second experiment.

(ii) Explain the shape and intercept of the graph you drew in (b)(i).[3]

**Answer/Explanation**

## Markscheme

a. kg m^{–1 }s^{–2 }K^{–1}

i any straight line that either goes or would go, if extended, through the origin

ii

for ideal gas *p* is proportional to *T* / P= nRT/V

gradient is constant /graph is a straight line

line passes through origin / 0,0

## 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.

a. 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.[1]

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.[3]

^{–3}\(\,\)m

^{2}and the temperature of air is 300 K. Estimate the number of moles of air in the tube.[2]

**Answer/Explanation**

## Markscheme

a. 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 × 10^{4} to 5.3 × 10^{4 }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]**

## Examiners report

a. In a simple pendulum experiment, a student measures the period *T* of the pendulum many times and obtains an average value *T* = (2.540 ± 0.005) s. The length *L* of the pendulum is measured to be *L* = (1.60 ± 0.01) m.

Calculate, using \(g = \frac{{4{\pi ^2}L}}{{{T^2}}}\), the value of the acceleration of free fall, including its uncertainty. State the value of the uncertainty to one significant figure. [3]

b. In a different experiment a student investigates the dependence of the period *T* of a simple pendulum on the amplitude of oscillations *θ*. The graph shows the variation of \(\frac{T}{{{T_0}}}\) with *θ*, where *T*_{0} is the period for small amplitude oscillations.

The period may be considered to be independent of the amplitude *θ* as long as \(\frac{{T – {T_0}}}{{{T_0}}} < 0.01\). Determine the maximum value of *θ* for which the period is independent of the amplitude. [2]

**Answer/Explanation**

### Markscheme

a. \(g = \frac{{4{\pi ^2} \times 1.60}}{{{{2.540}^2}}} = 9.7907\)

\(\Delta g = g\left( {\frac{{\Delta L}}{L} + 2 \times \frac{{\Delta T}}{T}} \right) = \) «\(9.7907\left( {\frac{{0.01}}{{1.60}} + 2 \times \frac{{0.005}}{{2.540}}} \right) = \)» 0.0997

**OR**

1.0%

hence g = (9.8 ± 0.1) «m\(\,\)s^{−2}» * OR* Δ

*g*= 0.1 «m\(\,\)s

^{−2}»

*For the first marking point answer must be given to at least 2 dp.**Accept calculations based on*

\({g_{\max }} = 9.8908\)

\({g_{\min }} = 9.6913\)

\(\frac{{{g_{\max }} – {g_{\min }}}}{2} = 0.099 \approx 0.1\)*[3 marks]*

*θ*_{max }= 22 «º»

*Accept answer from interval 20 to 24.**[2 marks]*

## Examiners report

A radio wave of wavelength \(\lambda \) is incident on a conductor. The graph shows the variation with wavelength \(\lambda \) of the maximum distance *d* travelled inside the conductor.

For \(\lambda \) = 5.0 x 10^{5} m, calculate the

The graph shows the variation with wavelength \(\lambda \) of *d *^{2}. Error bars are not shown and the line of best-fit has been drawn.

A student states that the equation of the line of best-fit is *d *^{2}^{ }= *a* + *b*\(\lambda \). When *d *^{2} and \(\lambda \) are expressed in terms of fundamental SI units, the student finds that *a* = 0.040 x 10^{–4} and *b* = 1.8 x 10^{–11}.

a. Suggest why it is unlikely that the relation between* d* and \(\lambda \) is linear. [1]

bi. fractional uncertainty in *d*.[2]

*d*

^{2}. [1]

*a*and of the constant

*b*.

2]

c.ii.Determine the distance travelled inside the conductor by very high frequency electromagnetic waves.[2]

**Answer/Explanation**

## Markscheme

a. it is not possible to draw a straight line through all the error bars**OR**

the line of best-fit is curved/not a straight line

*Treat as neutral any reference to the origin.*

*Allow “linear” for “straight line”. **[1 mark]*

*d*= 0.35 ± 0.01

*Δ*

**AND***d*= 0.05 ± 0.01 «cm»

«\(\frac{{\Delta d}}{d} = \frac{{0.5}}{{0.35}}\)» = 0.14

**OR**

\(\frac{1}{7}\) * or* 14%

*0.1*

**or***Allow final answers in the range of 0.11 to 0.18.*

*Allow [1 max] for 0.03 to 0.04 if \(\lambda \) = 5 × 10^{6} m is used.*

*[2 marks]**Allow ECF from (b)(i), but only accept answer as a %**[1 mark]*

*a:* m^{2}

*b:* m

*Allow answers in words**[2 marks]*

*– if graph on page 4 is used*

**ALTERNATIVE 1***d *^{2} = 0.040 x 10^{–4} «m^{2}»

*d* = 0.20 x 10^{–2} «m»

* ALTERNATIVE 2* – if graph on page 2 is used

any evidence that *d* intercept has been determined

*d* = 0.20 ± 0.05 «cm»

*For MP1 accept answers in range of 0.020 to 0.060 «cm ^{2}» if they fail to use given value of “a”.*

*For MP2 accept answers in range 0.14 to 0.25 «cm» .***[2 marks]**

A magnetized needle is oscillating on a string about a vertical axis in a horizontal magneticfield *B*. The time for 10 oscillations is recorded for different values of *B*.

The graph shows the variation with *B *of the time for 10 oscillations together with the uncertainties in the time measurements. The uncertainty in *B *is negligible.

a. Draw on the graph the line of best fit for the data. [1]

*B*= 0.005 T with its absolute uncertainty.[1]

*P*is given by:

\[P = \frac{K}{{\sqrt B }}\] where *K *is a constant.

Determine the value of *K *using the point for which *B *= 0.005 T.

State the uncertainty in *K *to an appropriate number of significant figures.[3]

*K*. [1]

*P*

^{2}varies with \(\frac{1}{B}\) for the data.

Sketch the shape of the expected line of best fit on the axes below assuming that the relationship \(P = \frac{K}{{\sqrt B }}\) is verified. You do **not **have to put numbers on the axes.

[2]

d. State how the value of *K *can be obtained from the graph.[1]

**Answer/Explanation**

## Markscheme

a. smooth line, not kinked, passing through all the error bars.**[1 mark]**

**«**s

**»**

*Accept any value from the range: 0.81 to 0.87.*

*Accept uncertainty 0.03 **OR **0.025. **[1 mark]*

**«**\(\frac{{\Delta K}}{K} = \frac{{\Delta P}}{P}\)**»**

\(\Delta K = \frac{{0.03}}{{0.84}} \times 0.0594 = 0.002\)

**«***K =*(0.059 ± 0.002)**»**

uncertainty given to 1sf

*Allow ECF **[3 max] **if 10T is used. **Award **[3] **for BCA.**[3 marks]*

*Accept *\(s\sqrt T \) or in words. **[1 mark]**

**ascending line through origin**

*AND*

**[2 marks]**

**[1 mark]**## Examiners report

A student carries out an experiment to determine the variation of intensity of the light with distance from a point light source. The light source is at the centre of a transparent spherical cover of radius *C*. The student measures the distance *x *from the surface of the cover to a sensor that measures the intensity *I *of the light.

The light source emits radiation with a constant power *P *and all of this radiation is transmitted through the cover. The relationship between *I *and *x *is given by

\[I = \frac{P}{{4\pi {{(C + x)}^2}}}\]

The student obtains a set of data and uses this to plot a graph of the variation of \(\frac{1}{{\sqrt I }}\) with *x*.

a. This relationship can also be written as follows.

\[\frac{1}{{\sqrt I }} = Kx + KC\]

Show that \(K = 2\sqrt {\frac{\pi }{P}} \).[1]

*C*. [2]

*P*, to the correct number of significant figures including its unit.[4]

*I*versus \(\frac{1}{{{x^2}}}\) has for the analysis in (b)(i) and (b)(ii). [2]

**Answer/Explanation**

## Markscheme

a. combines the two equations to obtain result

**«**for example \(\frac{1}{I}\) = *K*^{2}(*C* + *x*)^{2} = \(\frac{{4\pi }}{P}\)(*C* + *x*)^{2}**»**

*OR*

reverse engineered solution – substitute *K* = \(2\sqrt {\frac{\pi }{P}} \) into \(\frac{1}{I}\) = *K*^{2}(*C* + *x*)^{2} to get *I* = \(\frac{P}{{4\pi {{(C + x)}^2}}}\)

*There are many ways to answer the question, look for a combination of two equations to obtain the third one ***[1 mark]**

*x*-axis / use of

*x*-intercept

*OR*

Use *C* = \(\frac{{y{\text{ – intercept}}}}{{{\text{gradient}}}}\)

*OR*

use of gradient and one point, correctly substituted in one of the formulae

accept answers between 3.0 and 4.5 **«**cm**»**

*Award **[1 max] **for negative answers **[2 marks]*

*ALTERNATIVE 1*

Evidence of finding gradient using two points on the line at least 10 cm apart

Gradient found in range: 115–135 ** or **1.15–1.35

Using *P* = \(\frac{{4\pi }}{{{K^2}}}\) to get value between 6.9 × 10^{–4} and 9.5 × 10^{–4} **«**W**»** and POT correct

Correct unit, W **and **answer to 1, 2 or 3 significant figures

*ALTERNATIVE 2*

Finds \(I\left( {\frac{1}{{{y^2}}}} \right)\) from use of one point (*x *and *y*) on the line with *x* > 6 cm and *C* from(b)(i)to use in *I* = \(\frac{P}{{4\pi {{(C + x)}^2}}}\) or \(\frac{1}{{\sqrt I }}\) = *Kx* + *KC*

Correct re-arrangementto get *P *between 6.9 × 10^{–4} and 9.5 × 10^{–4} **«**W**»** and POT correct

Correct unit, W **and** answer to 1, 2 or 3 significant figures

*Award **[3 max] **for an answer between 6.9 W and 9.5 W (POT penalized in 3rd marking point)*

*Alternative 2 is worth **[3 max] **[4 marks]*

this graph will be a curve / not be a straight line

more difficult to determine value of *K*

*OR*

more difficult to determine value of *C*

*OR*

suitable mathematical argument *OWTTE **[2 marks]*