December 2023 – Physicspupil

December 20, 2023December 20, 2023

202312201537 Solution to 1980-CE-PHY-II-13

$6\times 10^{-3}\,\mathrm{m^3}$ of a gas is contained in a vessel at $91^\circ\mathrm{C}$ and a pressure of $4\times 10^5\,\mathrm{Pa}$ . If the density of the gas at s.t.p. ( $0^\circ\mathrm{C}$ and $10^5\,\mathrm{Pa}$ ) is $1.2\,\mathrm{kg\, m^{-3}}$ , what is the mass of the gas?

A. $7.2\,\mathrm{g}$
B. $14.4\,\mathrm{g}$
C. $21.6\,\mathrm{g}$
D. $28.8\,\mathrm{g}$

Official answer: C

Roughwork.

$\textrm{\textbf{\scriptsize{CASE I}}}$ (closed vessel)

Assumed a closed vessel, its volume $V=\textrm{Const.}$ a constant. Then, that $6\times 10^{-3}\,\mathrm{m^3}$ of gas fully occupied the closed vessel, implies that the volume of the vessel is also

$V=6\times 10^{-3}\,\mathrm{m^3}$ .

Hence may we write

$\begin{aligned} \textrm{density }(\rho ) & = \frac{\textrm{mass }(m)}{\textrm{volume }(V)} \\ m & = \rho V \\ & = (1.2\,\mathrm{kg\,m^{-3}})(6\times 10^{-3}\,\mathrm{m^3}) \\ & = 7.2\,\mathrm{g} \\ \end{aligned}$

And we get option A, which is different from the official answer C. Hence, we need to look into

$\textrm{\textbf{\scriptsize{CASE II}}}$ (open vessel)

We set up the background below.

$******************$

1. Ideal gas law:

An ideal gas satisfies the general gas equation

$pV=nRT$

where $p$ , $V$ , $n$ , $R$ , and $T$ are the pressure, the volume, the amount of substance (/number of moles), the ideal (/universal) gas constant, and the temperature of the gas.

2. Conversion of scales of temperature:

$x\,^\circ\mathrm{C} \approx (x+273)\,\mathrm{K}$

3. Relationship between chemical amount (/number of moles) $n$ , total mass $m$ , and molar mass $M$ , of a substance.

$\displaystyle{n=\frac{m}{M}}$

Note that total mass $m$ and number of moles $n$ are extensive (/extrinsic) properties; whereas molar mass $M$ is an intensive (/intrinsic) property.

$******************$

At the start $t=t_0$ ,

$\begin{aligned} p(t_0) & = 4\times 10^5\,\mathrm{Pa} \\ V(t_0) & = 6\times 10^{-3}\,\mathrm{m^3} \\ T(t_0) & = 364\,\mathrm{K} \\ R & = 8.31\mathrm{m^3\,Pa\,K^{-1}\,mol^{-1}} \\ \end{aligned}$

substituting,

$\begin{aligned} p(t_0)V(t_0) & = n(t_0)RT(t_0) \\ (4\times 10^5)(6\times 10^{-3}) & = n(t_0)(8.31)(364) \\ n(t_0) & = 0.7722\,\mathrm{mol}\quad\mathrm{(4\, s.f.)}\\ \end{aligned}$

at some point later $t=t_{1}$ the gas in standard temperature and pressure (s.t.p.),

$\begin{aligned} p(t_1) & = 10^5\,\mathrm{Pa} \\ V(t_1) & = 6\times 10^{-3}\,\mathrm{m^3} \\ T(t_1) & = 273\,\mathrm{K} \\ R & = 8.31\mathrm{m^3\,Pa\,K^{-1}\,mol^{-1}} \\ \end{aligned}$

substituting,

$\begin{aligned} p(t_1)V(t_1) & = n(t_1)RT(t_1) \\ (10^5)(6\times 10^{-3}) & = n(t_1)(8.31)(273) \\ n(t_1) & = 0.2645\,\mathrm{mol}\quad\mathrm{(4\, s.f.)}\\ \end{aligned}$

From $n(t_1)<n(t_0)$ , it can be inferred that some portion of the gas was leaking from the open vessel to the atmosphere throughout the experiment.

$\begin{aligned} \frac{m(t_1)}{n(t_1)} & = \frac{m(t_0)}{n(t_0)} = M \\ m(t_1) & = \bigg(\frac{n(t_1)}{n(t_0)}\bigg) m(t_0) \\ & = \bigg(\frac{0.2645}{0.7722}\bigg) m(t_0) \\ m(t_1) & = 0.3425m(t_0) \\ \end{aligned}$

Provided $\rho (t_1)=1.2\,\mathrm{kg\,m^{-3}}$ , we know

$\begin{aligned} \rho (t_1) & = \frac{m(t_1)}{V(t_1)} \\ \rho (t_1) & = \frac{Mn(t_1)}{V(t_1)} \\ 1.2 & = \frac{0.2645M}{6\times 10^{-3}} \\ M & = 0.0272212\,\mathrm{kg\,mol^{-1}}\\ \end{aligned}$

Hence

$\begin{aligned} m(t_1) & =Mn(t_1) \\ & = (0.0272212)(0.2645) \\ & = 0.0072000074\,\mathrm{kg} \\ & = 7.2\,\mathrm{g} \\ \end{aligned}$

And we get option A again, which is far from the official answer C.

Neither $\textrm{\textbf{\scriptsize{CASE I}}}$ nor $\textrm{\textbf{\scriptsize{CASE II}}}$ gives option C; have I actually made a circular argument?

Circular reasoning is a logical fallacy in which the reasoner begins with what they are trying to end with.

_{Wikipedia on Circular reasoning}

Have I really?

No, just that you do so in the way than intended.

December 19, 2023December 19, 2023

202312191041 Electrostatics Diagrams (Elementary) Q1

The blogger claims no originality of his problem below.

We have an arrangement of three positive (point) charges $+Q_{A}$ , $+Q_{B}$ , and $+Q_{C}$ of the same quantity ( $=Q$ ), as if lying in the vertices $A$ , $B$ , and $C$ of an equilateral triangle $\triangle ABC$ , where $\mathbf{r}_{BA}$ , $\mathbf{r}_{AC}$ , and $\mathbf{r}_{CB}$ are the same magnitude $r$ .

Intuitively, these three like charges are $\textrm{\scriptsize{NOT}}$ in electrostatic equilibrium for acting between them are $\textrm{\scriptsize{NET}}$ repulsive forces, not until we place a negative (point) charge $-q$ of some yet unknown quantity $q$ in the centre $O$ , pointing to which some attractive forces of yet also unknown magnitudes.

Express $q$ in terms of $Q$ and $r$ .

Setup.

$\begin{aligned} |\mathbf{r}_{BA}| & = |\mathbf{r}_{AC}| = |\mathbf{r}_{CB}| = r \\ |\mathbf{r}'_{A}| & = |\mathbf{r}'_{B}| = |\mathbf{r}'_{C}| = \frac{\sqrt{3}}{3}r \\ \mathbf{r}_{BA} & = \mathbf{r}'_{A} - \mathbf{r}'_{B} \\ \mathbf{r}_{AC} & = \mathbf{r}'_{C} - \mathbf{r}'_{A} \\ \mathbf{r}_{CB} & = \mathbf{r}'_{B} - \mathbf{r}'_{C} \\ \text{}_{B}\mathbf{F}_{A} & = \text{}_{B}F_{A}\,\hat{\mathbf{r}}_{BA} \\ \text{}_{C}\mathbf{F}_{A} & = -\text{}_{C}F_{A}\,\hat{\mathbf{r}}_{AC} \\ \text{}_{C}\mathbf{F}_{B} & = \text{}_{C}F_{B}\,\hat{\mathbf{r}}_{CB} \\ \text{}_{A}\mathbf{F}_{B} & = -\text{}_{A}F_{B}\,\hat{\mathbf{r}}_{BA} \\ \text{}_{A}\mathbf{F}_{C} & = \text{}_{A}F_{C}\,\hat{\mathbf{r}}_{AC} \\ \text{}_{B}\mathbf{F}_{C} & = -\text{}_{B}F_{C}\,\hat{\mathbf{r}}_{CB} \\ \text{}_{B}F_{A} & = |\text{}_{B}\mathbf{F}_{A}| = F \\ \text{}_{C}F_{A} & = |\text{}_{C}\mathbf{F}_{A}| = F \\ \text{}_{C}F_{B} & = |\text{}_{C}\mathbf{F}_{B}| = F \\ \text{}_{A}F_{B} & = |\text{}_{A}\mathbf{F}_{B}| = F \\ \text{}_{A}F_{C} & = |\text{}_{A}\mathbf{F}_{C}| = F \\ \text{}_{B}F_{C} & = |\text{}_{B}\mathbf{F}_{C}| = F \\ \mathbf{F}_{A} & = \text{}_{B}\mathbf{F}_{A} + \text{}_{C}\mathbf{F}_{A} \\ \mathbf{F}_{B} & = \text{}_{C}\mathbf{F}_{B} + \text{}_{A}\mathbf{F}_{B} \\ \mathbf{F}_{C} & = \text{}_{A}\mathbf{F}_{C} + \text{}_{B}\mathbf{F}_{C} \\ \mathbf{F}_{A} & = F_{A}\,\hat{\mathbf{r}}'_{A} \\ \mathbf{F}_{B} & = F_{B}\,\hat{\mathbf{r}}'_{B} \\ \mathbf{F}_{C} & = F_{C}\,\hat{\mathbf{r}}'_{C} \\ F_{A} & = |\mathbf{F}_{A}| = F' \\ F_{B} & = |\mathbf{F}_{B}| = F' \\ F_{C} & = |\mathbf{F}_{C}| = F' \\ \end{aligned}$

Write, by Coulomb’s law,

$\begin{aligned} F & = k\frac{(Q)(Q)}{(r)^2} \\ & = k\frac{Q^2}{r^2} \\ \end{aligned}$

such that

$\begin{aligned} \text{}_{B}\mathbf{F}_{A} & = F\cos 60^\circ\,\hat{\mathbf{i}} + F\sin 60^\circ\,\hat{\mathbf{j}} \\ & = \bigg( k\frac{Q^2}{r^2}\bigg) \bigg(\frac{1}{2}\bigg) \,\hat{\mathbf{i}} + \bigg( k\frac{Q^2}{r^2}\bigg) \bigg(\frac{\sqrt{3}}{2}\bigg) \,\hat{\mathbf{j}} \\ \text{}_{C}\mathbf{F}_{A} & = -F\cos 60^\circ\,\hat{\mathbf{i}} + F\sin 60^\circ\,\hat{\mathbf{j}} \\ & = -\bigg( k\frac{Q^2}{r^2}\bigg) \bigg(\frac{1}{2}\bigg) \,\hat{\mathbf{i}} + \bigg( k\frac{Q^2}{r^2}\bigg) \bigg(\frac{\sqrt{3}}{2}\bigg) \,\hat{\mathbf{j}} \\ \mathbf{F}_{A} & = \text{}_{B}\mathbf{F}_{A} + \text{}_{C}\mathbf{F}_{A} \\ & = 0\,\hat{\mathbf{i}} + \bigg( k\frac{Q^2}{r^2}\bigg)(\sqrt{3})\,\hat{\mathbf{j}} \\ F' & = |\mathbf{F}_A| \\ & = \bigg( k\frac{Q^2}{r^2}\bigg)(\sqrt{3}) \\ \end{aligned}$

and

$\begin{aligned} F' & = k\frac{(Q)(q)}{(\sqrt{3}r/3)^2} \\ \bigg( k\frac{Q^2}{r^2}\bigg) (\sqrt{3}) & = \bigg( k\frac{Qq}{r^2}\bigg) (3)\\ q & = \frac{\sqrt{3}}{3} Q \\ \end{aligned}$

$\therefore$ We should put a negative charge $-q$ of quantity $\sqrt{3}Q/3$ in order for the system to reach electrostatic equilibrium.

This problem is not to be attempted.

December 18, 2023December 18, 2023

202312181044 Exercise 21.2.27

The figure below shows two masses connected by a light inextensible string on a smooth incline where $\sin\theta =\frac{1}{40}$ . Find the acceleration of the masses when the system is released from rest. ( $g=9.8$ )

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

As always, begin with free-body diagrams. Hence, draw

$\begin{aligned} m_Aa_A & =\textrm{Net }F_{A}=-T+W_{A}\sin\theta \\ m_Ba_B & =\textrm{Net }F_{B}=T+W_{B}\sin\theta \\ \end{aligned}$

where

$\begin{aligned} W_A & = m_Ag \\ W_B & = m_Bg \\ T & : \left\{\begin{array}{lr} =0 \textrm{ for }a_A\leqslant a_B & \text{} \\ >0 \textrm{ for }a_A>a_B & \text{}\\ \end{array} \\ \end{aligned}$

Note that

$\begin{aligned} a_A & = -\frac{T}{m_A}+g\sin\theta \\ a_B & = \frac{T}{m_B}+g\sin\theta \\ \cdots\cdots & \cdots\cdots\cdots \\ \because\quad a_A & \leqslant a_B \\ \therefore\quad T & = 0 \\ \end{aligned}$

Thus,

$a_A,a_B = g\sin\theta$ .

This problem is not to be attempted.