Съдържание

Instructions Table of Constants Условие (a) (2 т.)(b) Criticality (6 т.)(c) The gas phase (4 т.)(d) The liquid phase (5 т.)Решение

INPhO 2018, Теория — Задача 6. The Van der Waals Gas

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от INPhO 2018

Indian National Physics Olympiad – 2018 · 28 януари 2018 г. · 17 т.

Внимание: Бележка към темата

All seven matching archived keys are retained, including alternative ladder methods, unequal-resistor generalization, accepted ranges and the full handwritten graph. Printed refraction, hourglass, Fibonacci and notation inconsistencies remain literal with separate source notes. The full original includes nineteen blank detailed-answer sheets and two graph papers.

Native cover and instructions

Indian National Physics Olympiad – 2018

Date: 28 January 2018. Time: 09:00–12:00 (3 hours). Maximum Marks: 75.

Roll Number: $1\,8\,\square\,\square-\square\,\square\,\square\,\square-\square\,\square\,\square\,\square$ .

I permit/do not permit (strike out one) HBCSE to reveal my academic performance and personal details to a third party.

Besides the International Physics Olympiad (IPhO) 2018, do you also want to be considered for the Asian Physics Olympiad (APhO) 2018? For APhO 2018 and its pre-departure training, your presence will be required in Delhi and Vietnam from April 28 to May 15, 2018. In principle, you can participate in both Olympiads. Yes/No.

Full Name (BLOCK letters) Ms./Mr.: [blank].

Extra sheets attached: [blank box]. Date: [blank]. Centre (e.g. Kochi): [blank]. Signature: [blank].

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Instructions

This booklet consists of 27 pages (excluding this sheet) and total of 7 questions.
This booklet is divided in two parts: Questions with Summary Answer Sheet and Detailed Answer Sheet. Write roll number at the top wherever asked.
The final answer to each sub-question should be neatly written in the box provided below each sub-question in the Questions & Summary Answer Sheet.
You are also required to show your detailed work for each question in a reasonably neat and coherent way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of these pages.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations. Strike out any rough work that you do not want to be considered for evaluation.
Adequate space has been provided in the answersheet for you to write/calculate your answers. In case you need extra space to write, you may request additional blank sheets from the invigilator. Write your roll number on the extra sheets and get them attached to your answersheet and indicate number of extra sheets attached at the top of this page.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Use blue or black pen to write answers. Pencil may be used for diagrams/graphs/sketches.
This entire booklet must be returned at the end of the examination.

Table of Constants

Quantity	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Planck’s constant	$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}$
	$\hbar$	$h/2\pi$
Universal constant of Gravitation	$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Rest mass of electron	$m_e$	$9.11\times10^{-31}\ \mathrm{kg}$
Rest mass of proton	$m_p$	$1.67\times10^{-27}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00\times10^9\ \mathrm{N\cdot m^2\cdot C^{-2}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
Molar mass of water		$18.02\ \mathrm{g\cdot mol^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Question	Marks	Score
1	4
2	8
3	12
4	9
5	10
6	17
7	15
Total	75

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Native key cover and marking policy

Solutions of Indian National Physics Olympiad – 2018

The key cover repeats the date, roll-number form, APhO participation question, nine instructions, thirteen constant rows, seven-question score table and HBCSE address of the question cover, but prints “16 pages (excluding this sheet)”. The retained key has eighteen PDF pages: the cover, sixteen numbered pages and one appended handwritten graph sheet.

On its next page the key states: ANY ALTERNATIVE METHOD OF SOLUTION TO ANY QUESTION THAT IS SCIENTIFICALLY AND MATHEMATICALLY CORRECT, AND LEADS TO THE SAME ANSWER WILL BE ACCEPTED WITH FULL CREDIT. PARTIALLY CORRECT ANSWERS WILL GAIN PARTIAL CREDIT.

Условие

Consider $n$ mole of a non-ideal (realistic) gas. Its equation of state maybe described by the Van der Waals equation

\left(P+\frac{an^2}{V^2}\right)\left(\frac Vn-b\right)=RT

where $a$ and $b$ are positive constants. We take one mole of the gas ( $n=1$ ). You must bear in mind that one is often required to make judicious approximations to understand realistic systems.

(a) (2 т.)

For this part only take $a=0$ . Obtain expressions in terms of $V$ , $T$ and constants for

i. the coefficient of volume expansion ( $\beta$ );

Summary answer field: $\beta=$ [blank]. [1 т.]

ii. the isothermal compressibility ( $\kappa$ ).

Summary answer field: $\kappa=$ [blank]. [1 т.]

(b) Criticality (6 т.)

The Van der Waals gas exhibits phase transition. A typical isotherm at low temperature is shown in the figure. Here $L$ ( $G$ ) represents the liquid (gas) phase and at $P_{LG}$ there are three possible solutions for the volume ( $V_L$ , $V_{LG}$ , $V_G$ ). As the temperature is raised, at a certain temperature $T_c$ , the three values of the volume merge to a single value, $V_c$ (corresponding pressure being $P_c$ ). This is called the point of criticality. As the temperature is raised further there exists only one real solution for the volume and the isotherm resembles that of an ideal gas.

Both original Van der Waals isotherm panels with full legend, L/LG/G, PLG/VL/VLG/VG labels and native captions.

i. Obtain the critical constants $P_c$ , $V_c$ and $T_c$ in terms of $a$ , $b$ and $R$ .

Summary answer fields: $P_c=$ [blank]; $V_c=$ [blank]; $T_c=$ [blank]. [4 т.]

ii. Obtain the values of $a$ and $b$ for $\mathrm{CO_2}$ given $T_c=3.04\times10^2\ \mathrm K$ and $P_c=7.30\times10^6\ \mathrm{N\cdot m^{-2}}$ .

Summary answer fields: $a=$ [blank]; $b=$ [blank]. [1 т.]

iii. The constant $b$ represents the volume of the gas molecules of the system. Estimate the size $d$ of a $\mathrm{CO_2}$ molecule.

Summary answer field: $d=$ [blank]. [1 т.]

(c) The gas phase (4 т.)

For the gaseous phase the volume $V_G\gg b$ . Let the pressure $P_{LG}=P_0$ , the saturated vapour pressure.

i. Obtain the expression for $V_G$ in terms of $R$ , $T$ , $P_0$ and $a$ .

Summary answer field: $V_G=$ [blank]. [1,5 т.]

ii. State the corresponding expression for $V_I$ for an ideal gas.

Summary answer field: $V_I=$ [blank]. [0,5 т.]

iii. Obtain $(V_G-V_I)/V_I$ for water given $T=1.00\times10^2\ {}^\circ\mathrm C$ , $P_0=1.00\times10^5\ \mathrm{Pa}$ , $b=3.10\times10^{-5}\ \mathrm{m^3\cdot mol^{-1}}$ and $a=0.56\ \mathrm{m^6\cdot Pa\cdot mol^{-2}}$ . Comment on your result.

Summary answer fields: $(V_G-V_I)/V_I=$ [blank]; Comment: [blank]. [2 т.]

(d) The liquid phase (5 т.)

For the liquid phase $P\ll a/V_L^2$ .

i. Obtain the expression for $V_L$ .

Summary answer field: $V_L=$ [blank]. [1,5 т.]

ii. Obtain the density of water ( $\rho_w$ ). You may take the molar mass to be $1.80\times10^{-2}\ \mathrm{kg\cdot mole^{-1}}$ .

Summary answer field: $\rho_w=$ [blank]. [1,5 т.]

iii. The heat of vaporization is the energy required to overcome the attractive intermolecular force as the system is taken from the liquid phase ( $V_L$ ) to the gaseous phase ( $V_G$ ). The term $a/V^2$ represents this. Obtain the expression for the specific heat of vaporization per unit mass ( $L$ ) and obtain its value for water.

Summary answer fields: $L=$ [blank]; Value of $L=$ [blank]. [2 т.]

Detailed answers can be found on page numbers: [blank].

Решение

(a) i.

\beta=\frac1V\left(\frac{dV}{dT}\right)_p=\frac{V-b}{VT}\qquad (22)

(a) ii.

\kappa=-\frac1V\left(\frac{dV}{dP}\right)_T=\frac{(V-b)^2}{VRT}\qquad (23)

(b) i.

At criticality, the curve is a cubic like function. Hence

\left(\frac{dP}{dV}\right)_T=\left(\frac{d^2P}{dV^2}\right)_T=0

Above condition along with Van der Waal’s equation can be used to obtain

V_C=3b\qquad (24)

T_C=\frac{8a}{27Rb}\qquad (25)

P_C=\frac a{27b^2}\qquad (26)

(b) ii.

$a=0.37\ \mathrm{m^6\cdot Pa/mol^2}$ and $b=4.33\times10^{-5}\ \mathrm{m^3/mol}$ .

(b) iii.

b=N_Ad^3\quad\text{or}\quad N_A4\pi r^3/3

d=1.0\times10^{-10}\ \mathrm m\ \text{to}\ 6.0\times10^{-10}\ \mathrm m

(c) i.

P_0V_G^2-RTV_G+a=0

V_G=\frac{RT}{2P_0}\left(1\pm\left(1-\frac{4aP_0}{R^2T^2}\right)^{1/2}\right)

Taking positive sign since in the ideal gas limit $a\longrightarrow0$ , $V_I=RT/P_0$ for one mole.

V_G\simeq\frac{RT}{P_0}-\frac a{RT}

(c) ii.

V_I=\frac{RT}{P_0}

(c) iii.

Comment 1: For ideal gas, $a/RT$ vanishes as $T\to\infty$ .

\frac{V_G-V_I}{V_I}=\frac{-aP_0}{(RT)^2}\simeq-5.50\times10^{-3}\ \text{to}\ -7.00\times10^{-3}

Comment 2: Answer is negative, indicating attractive intermolecular forces.

(d) i.

In this phase $P\ll a/V_L^2$ . Hence

\frac a{V_L^2}(V_L-b)=RT

V_L=\frac a{2RT}\left(1\pm\sqrt{1-\frac{4bRT}a}\right)

Taking negative sign since as $T\to0$ , $V_L\to b$ .

\begin{aligned}V_L&=\frac a{2RT}\left(1-\sqrt{1-\frac{4bRT}a}\right)\\&\simeq b\left(1+\frac{bRT}a\right)\end{aligned}

(d) ii.

\rho_w\approx\frac{1.80\times10^{-2}}b=3.60\times10^2\ \text{to}\ 6.00\times10^2\ \mathrm{Kg/m^3}\qquad (27)

(d) iii.

L\times\text{molar mass}=\int_{V_L}^{V_G}\frac a{V^2}\,dV\simeq\frac a{V_L}

L\simeq6.0\times10^5\ \text{to}\ 1.1\times10^6\ \mathrm{J/Kg}

Source note: The density and heat-of-vaporization expressions are the approximations printed in the key. Its accepted ranges are retained in full; the source prints capital “Kg” in the two unit labels and upper-case critical subscripts $C$ in Eqs. (24)–(26).

Оригинал в Архива: INPhO2018-Question.pdf · решение от архива: INPhO2018-Solution-20180213.pdf