| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Laws of Logarithms |
| Type | Model y=ax^b: linearise and find constants from graph/data |
| Difficulty | Moderate -0.3 This is a standard logarithmic transformation question requiring reading coordinates from a graph, finding gradient and intercept, then using the model for prediction. While it involves multiple steps (8 marks total), each step follows routine A-level procedures: reading graph coordinates, calculating gradient from two points, finding the y-intercept, and substituting into the model. The conceptual demand is modest—students who understand log-linear graphs can follow the mechanical steps without novel insight. |
| Spec | 1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| 8(a) | Reads graph and uses 10logP or | |
| 10logV to get either P or V | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWRT 150 and AWRT 0.71 | AO1.1b | A1 |
| (b) | Calculates value of gradient to find d |
| Answer | Marks | Guidance |
|---|---|---|
| to find d | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Not necessarily a decimal | AO1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| to find c | AO3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWRT 93 | AO1.1b | A1 |
| (c) | Uses their values of c and d in the | |
| formula P = cVd | AO1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| AWFW 30 to 40 | AO3.2a | A1 |
| Total | 8 | |
| Q | Marking Instructions | AO |
Question 8:
--- 8(a) ---
8(a) | Reads graph and uses 10logP or
10logV to get either P or V | AO3.4 | M1 | log P = 2.18
10
P = 151
log V = –0.15
10
V = 0.708
Correctly obtains P and V
AWRT 150 and AWRT 0.71 | AO1.1b | A1
(b) | Calculates value of gradient to find d
Condone use of log d = gradient
Or uses a value of c plus a P/V pair
to find d | AO3.4 | M1 | log P = log c + d log V
10 10 10
Gradient = d = –1.4
Intercept = log c
10
c = 93.3
Obtains correct value for d
AWRT –1.4
Not necessarily a decimal | AO1.1b | A1
Calculates value of intercept to find
log c
10
Or uses a value of d plus a P/V pair
to find c | AO3.4 | M1
Calculates correct value for c
AWRT 93 | AO1.1b | A1
(c) | Uses their values of c and d in the
formula P = cVd | AO1.1a | M1 | P = 93.3 × 2–1.4
= 35.4 kilopascals
Obtains P value, including units
AWFW 30 to 40 | AO3.2a | A1
Total | 8
Q | Marking Instructions | AO | Marks | Typical SolutionMaxine measures the pressure, $P$ kilopascals, and the volume, $V$ litres, in a fixed quantity of gas.
Maxine believes that the pressure and volume are connected by the equation
$$P = cV^d$$
where $c$ and $d$ are constants.
Using four experimental results, Maxine plots $\log_{10} P$ against $\log_{10} V$, as shown in the graph below.
\includegraphics{figure_8}
\begin{enumerate}[label=(\alph*)]
\item Find the value of $P$ and the value of $V$ for the data point labelled $A$ on the graph.
[2 marks]
\item Calculate the value of each of the constants $c$ and $d$.
[4 marks]
\item Estimate the pressure of the gas when the volume is $2$ litres.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2018 Q8 [8]}}