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INPhO 2025, Теория — Задача 3. Love is in the air

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Indian National Physics Olympiad (INPhO)-2025 — Question Paper · 2 февруари 2025 г. · 16 т.

Внимание: Бележка към темата

The complete retained original question and matching HBCSE solution packets are transcribed. The separately referenced Answer Booklet is not retained in the matching archive row. Printed key formula/sign/symbol/prose discrepancies and the original handwritten magnetic-data graph remain literal, with separately labelled source notes.

Original cover and instructions

Indian National Physics Olympiad (INPhO)-2025

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

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

Question Paper

Date: 02 February 2025

Time: 09:00-12:00 (3 hours)

Maximum Marks: 75

Instructions — Roll Number: $\square\square\square\square-\square\square\square\square-\square\square\square\square$

This booklet consists of 9 pages and total of 5 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.

Question Number	1	2	3	4	5	Total
Maximum Marks	8	14	16	22	15	75

Table of Constants

Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
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	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Boltzmann constant	$k_B$	$1.3806\times10^{-23}\ \mathrm{J\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Native key notice

The matching key repeats the cover, constant table, marks table and four instructions with “24 pages” and the title “Solutions”.

Please note that alternate/equivalent methods and different ways of expressing final solutions may exist. A correct method will be suitably awarded.

Условие

A thermodynamic cycle is performed for one mole of an ideal monoatomic gas. The representation of this cycle is in the shape of a “heart” in the volume $(V)$ – temperature $(T)$ graph (shown as the shaded area below). However, neither of the axes are provided in the graph. The graph is drawn to scale with 1 cm along the $V$ -axis representing 4 litre, and 1 cm along the $T$ -axis representing 80 K.

It is given that the pressure at the point $X$ is minimum for the whole cycle, with the temperature and volume at this point being $T_X=224\ \mathrm K$ and $V_X=24\ \text{litre}$ .

Complete original rectangular scale/drawing box with whole shaded heart curve and X≡(24L,224K) native red label.

a) Draw both the $V$ and $T$ axes to scale in the same diagram given in the Summary Answersheet. Indicate the origin by “O”. Justify your answer in the detailed answersheet. You are given one extra answer box in the answersheet, in case of any mistake in the first. [13 т.]

b) For the axes and origin you have drawn, indicate the point(s) on the graph where the pressure is/are maximum in the cycle by $\otimes$ and label it as $P_{\max}$ on the curve. Determine the value of the maximum pressure. [3 т.]

Решение

(a) Solution: We begin with the ideal gas law:

PV=nRT\qquad(3.1)

Differentiating with respect to volume, $V$ , we get:

P+V\frac{dP}{dV}=nR\frac{dT}{dV}\qquad(3.2)

\frac{RT}{V}+V\frac{dP}{dV}=nR\frac{dT}{dV}\qquad(3.3)

T=n\frac{dT}{dV}V-\frac{V^2}{R}\frac{dP}{dV}\qquad(3.4)

At points of extreme pressure (maxima or minima), we have $\frac{dP}{dV}=0$ . Equation (3.4) then simplifies to:

T=n\frac{dT}{dV}V\qquad(3.5)

This implies that the tangent drawn to the $T$ - $V$ curve at these extreme pressure points must pass through the origin of the $T$ - $V$ graph. Let’s denote the origin as point O. Point X represents a pressure minimum (as given), and we need to find points where pressure is maximum.

We are given the volume $V_X=24\ \mathrm L$ , which scales to 6 cm on the graph, and the temperature $T_X=224\ \mathrm K$ , which scales to 2.8 cm. Using these scaled values, the distance OX can be calculated as:

OX=\sqrt{V_X^2+T_X^2}=\sqrt{6^2+2.8^2}\approx6.6\ \mathrm{cm}\qquad(3.6)

From Equation (3.4), for $n=1$ , at point X, we have

\left.\frac{dT}{dV}\right|_X=\frac{T_X}{V_X}\qquad(3.7)

The slope of the tangent at X with respect to the volume axis is given by:

\frac{dT}{dV}=\tan(\theta)=\frac{T_X}{V_X}=\frac{2.8}{6}

Thus, the angle the tangent makes with the volume axis, $\theta$ , is:

\theta=\arctan\left(\frac{2.8}{6}\right)\approx25^\circ

To draw the axes:

Draw a tangent line to the curve at point X.
Locate the origin (O) at a distance of 6.6 cm from point X on the tangent line. This means there are two possible locations for the origin, one on either side of X along the tangent.
Draw the volume (V) axis by drawing a line passing through O that makes an angle of approximately 25° with the tangent line at X.
Draw the temperature (T) axis by drawing a line perpendicular to the V axis and passing through the origin (O).

Note that the axes are oriented such that no part of the cycle falls into the negative T or V quadrants.

We therefore have two possible sets of axes, corresponding to the two possible origins. The solutions are displayed below, and either answer will be accepted. Origins are marked here by $O_1$ and $O_2$ .

Full native boxed heart graph with O1, both blue tangents, rotated V/T axes, X, 25-degree angle and circled-cross Pmax.

Full second native boxed heart graph with O2, both blue tangents, rotated V/T axes, X, 25-degree angle and circled-cross Pmax.

(b) Solution: Once the axes are drawn, our next goal is to identify the point, which we call $P_{\max}$ , where the slope of the tangent line to the curve is maximized and this tangent line passes through the origin. In other words, we seek the point on the $T$ - $V$ curve with the steepest tangent that extends back to the origin.

Having located point $P_{\max}$ , we can then determine its coordinates in “cm” as measured on our defined axes. Multiplying these coordinate values by the corresponding scale factors for temperature and volume will give the volume and temperature in L and K respectively. Then we can use ideal gas law to obtain the maximum pressure on these points.

The point $P_{\max}$ where the pressure is maximum is indicated in both the possible axes. Exact numerical values for these parameters are provided in the table below. Answers within a reasonable tolerance will be considered acceptable.

Origin	Coordinates $(V,T)$ : $V$	Coordinates $(V,T)$ : $T$	Values $(V,T)$ : $V$	Values $(V,T)$ : $T$	Pressure value (atm)
$O_1$	4.8 cm	8.0 cm	19.2 L	640.0 K	2.7 (accetable range is 2.4 to 2.8)
$O_2$	1.3 cm	5.4 cm	5.2 L	432.0 K	6.8 (accetable range is 6.3 to 7.5)

Source note: Equation (3.3) prints $RT/V$ without $n$ , while (3.4)–(3.5) place $n$ as shown. Equation (3.6) uses the physical symbols for scaled centimetre coordinates. The pressure-table wording “accetable” and the prose ordering “temperature and volume ... volume and temperature” are literal. Both accepted native axis constructions are retained.

Native source end

END OF THE QUESTION PAPER

The key also ends with this native wording on page24. Key page20 is blank apart from running page and roll-number furniture.

Оригинал в Архива: INPhO2025-Question.pdf · официални решения: INPhO2025-Solution-20250219.pdf