December 20, 2023 – Physicspupil

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

Day: December 20, 2023

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