| Exam Board | OCR |
|---|---|
| Module | Further Mechanics (Further Mechanics) |
| Year | 2018 |
| Session | December |
| Marks | 17 |
| Topic | Dimensional Analysis |
| Type | Derive dimensions from formula |
| Difficulty | Standard +0.3 This is a straightforward dimensional analysis question requiring systematic application of standard techniques: deriving dimensions from definitions (force/area for pressure), using given units (J = ML²T⁻²), and applying dimensional homogeneity. While multi-part with several steps, each part follows directly from the previous with no novel insight required—slightly easier than average A-level work. |
| Spec | 6.01a Dimensions: M, L, T notation6.01b Units vs dimensions: relationship6.01c Dimensional analysis: error checking6.01d Unknown indices: using dimensions |
| Answer | Marks | Guidance |
|---|---|---|
| \([F] = [ma] = MLT^{-2}\) | M1 [1.2] | |
| \([P] = \frac{[F]}{[A]} = MLT^{-2}L^{-2} = ML^{-1}T^{-2}\) | A1 [2.2a] | |
| \([R] = [F][d]\) | M1 [3.1b] | Determining dimensions of energy |
| \([R] = MLT^{-2}L = ML^2T^{-2}\) | A1 [3.2a] | |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \((ML^{-1}T^{-2})^{\alpha}L^3\beta = ML^2T^{-2}\) | M1 [3.3] | Using their dimensions |
| \(M^{\alpha} = M \Rightarrow \alpha = 1\) (AG) | A1 [3.4] | or similarly with T |
| \(L: -\alpha + 3\beta = 2 \Rightarrow \beta = 1\) (AG) | A1 [1.1] | |
| [3] | \(n\) and \(\theta\) dimensionless |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = \frac{5 \times 8.31 \times 300}{0.03}\) | M1 [3.4] | Correctly substituting values |
| 415 500 N m\(^{-2}\) | A1 [1.1] | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \([b] = L^3\) | B1 [2.2a] | |
| \([a] = ML^{-1}T^{-2}(L)^2\) | M1 [2.2a] | Using equality of \([a/V^2]\) and \([P]\) |
| \([a] = ML^2T^{-2}\) | A1 [1.1] | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(p + \frac{0.14 \times 5^2}{0.03^2} = \frac{5 \times 8.31 \times 300}{0.03 - 5 \times 3.2 \times 10^{-5}}\) | M1 [3.4] | Correctly substituting values |
| awrt 414 000 N m\(^{-2}\) | A1 [1.1] | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(P = \frac{5 \times 8.31 \times 300}{1.5 \times 10^{-4} - 5 \times 3.2 \times 10^{-5}} - \frac{0.14 \times 5^2}{(1.5 \times 10^{-4})^2} \approx -1.4 \times 10^9\) | B1 [3.4] | Ignore units |
| A negative value is not a valid value for pressure... ...and so the model is not valid for these values | E1 [3.5b], E1 [2.2a] | |
| [3] | OR from \(V - nb < 0\), so since \(R\), \(\theta\), \(n\) and \(V\) are all positive \(P < 0\) |
### (a)
$[F] = [ma] = MLT^{-2}$ | M1 [1.2]
$[P] = \frac{[F]}{[A]} = MLT^{-2}L^{-2} = ML^{-1}T^{-2}$ | A1 [2.2a]
$[R] = [F][d]$ | M1 [3.1b] | Determining dimensions of energy
$[R] = MLT^{-2}L = ML^2T^{-2}$ | A1 [3.2a]
| [4]
### (b)
$(ML^{-1}T^{-2})^{\alpha}L^3\beta = ML^2T^{-2}$ | M1 [3.3] | Using their dimensions
$M^{\alpha} = M \Rightarrow \alpha = 1$ (AG) | A1 [3.4] | or similarly with T
$L: -\alpha + 3\beta = 2 \Rightarrow \beta = 1$ (AG) | A1 [1.1]
| [3] | $n$ and $\theta$ dimensionless
### (c)
$p = \frac{5 \times 8.31 \times 300}{0.03}$ | M1 [3.4] | Correctly substituting values | With their $\alpha$ and $\beta$
415 500 N m$^{-2}$ | A1 [1.1]
| [2]
### (d)
$[b] = L^3$ | B1 [2.2a]
$[a] = ML^{-1}T^{-2}(L)^2$ | M1 [2.2a] | Using equality of $[a/V^2]$ and $[P]$ | With $n$ dimensionless
$[a] = ML^2T^{-2}$ | A1 [1.1]
| [3]
### (e)
$p + \frac{0.14 \times 5^2}{0.03^2} = \frac{5 \times 8.31 \times 300}{0.03 - 5 \times 3.2 \times 10^{-5}}$ | M1 [3.4] | Correctly substituting values | With their $\alpha$ and $\beta$
awrt 414 000 N m$^{-2}$ | A1 [1.1]
| [2]
### (f)
$P = \frac{5 \times 8.31 \times 300}{1.5 \times 10^{-4} - 5 \times 3.2 \times 10^{-5}} - \frac{0.14 \times 5^2}{(1.5 \times 10^{-4})^2} \approx -1.4 \times 10^9$ | B1 [3.4] | Ignore units
A negative value is not a valid value for pressure... ...and so the model is not valid for these values | E1 [3.5b], E1 [2.2a]
| [3] | OR from $V - nb < 0$, so since $R$, $\theta$, $n$ and $V$ are all positive $P < 0$
---6 This question is about modelling the relation between the pressure, $P$, volume, $V$, and temperature, $\theta$, of a fixed amount of gas in a container whose volume can be varied.
The amount of gas is measured in moles; 1 mole is a dimensionless constant representing a fixed number of molecules of gas. Gas temperatures are measured on the Kelvin scale; the unit for temperature is denoted by K . You may assume that temperature is a dimensionless quantity.
A gas in a container will always exert an outwards force on the walls of the container. The pressure of the gas is defined to be the magnitude of this force per unit area of the walls, with $P$ always positive.
An initial model of the relation is given by $P ^ { \alpha } V ^ { \beta } = n R \theta$, where $n$ is the number of moles of gas present and $R$ is a quantity called the Universal Gas Constant. The value of $R$, correct to 3 significant figures, is $8.31 \mathrm { JK } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $[ P ] = \mathrm { ML } ^ { - 1 } \mathrm {~T} ^ { - 2 }$ and $[ R ] = \mathrm { ML } ^ { 2 } \mathrm {~T} ^ { - 2 }$.
\item Hence show that $\alpha = 1$ and $\beta = 1$.
5 moles of gas are present in the container which initially has volume $0.03 \mathrm {~m} ^ { 3 }$ and which is maintained at a temperature of 300 K .
\item Find the pressure of the gas, as predicted by the model.
An improved model of the relation is given by $\left( P + \frac { a n ^ { 2 } } { V ^ { 2 } } \right) ( V - n b ) = n R \theta$, where $a$ and $b$ are constants.
\item Determine the dimensions of $b$ and $a$.
The values of $a$ and $b$ (in appropriate units) are measured as being 0.14 and $3.2 \times 10 ^ { - 5 }$ respectively.
\item Find the pressure of the gas as predicted by the improved model.
Suppose that the volume of the container is now reduced to $1.5 \times 10 ^ { - 4 } \mathrm {~m} ^ { 3 }$ while keeping the temperature at 300 K .
\item By considering the value of the pressure of the gas as predicted by the improved model, comment on the validity of this model in this situation.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics 2018 Q6 [17]}}