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EuPhO 2021, Теория — Задача 1. A leak

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Всички материали от EuPhO 2021

EuPhO-2021 Theoretical Problems. Language: English

Условие

A hollow insulated cylinder of height $2H$ and volume $2V$ is closed from below by an insulating piston. The cylinder is divided into two initially identical chambers by an insulating diaphragm of mass $m$ . The diaphragm rests on a circular ledge and a gasket between them provides tight contact. Both chambers are filled with gaseous helium at pressure $p$ and temperature $T$ . A force is applied to the piston, so that it moves upwards slowly.

Vertical insulated cylinder of height 2H divided by a diaphragm of mass m into two chambers, each labelled p, V, T; a piston with a rod closes the cylinder from below; an arrow shows gravity g pointing down.

a. Find the volume of the lower chamber $V_0$ when the gas starts to leak between the chambers.

b. Find the temperature $T_1$ in the upper chamber when the piston touches the diaphragm.

c. Find the temperature $T_2$ in the lower chamber immediately before the piston touches the diaphragm.

Отговори

Решение

Let $p_1$ , $V_1$ , $T_1$ denote the (time dependent) pressure, volume, and temperature in the upper chamber, and $p_2$ , $V_2$ , $T_2$ — those in the lower one. Note that $V_1 \equiv V$ does not change.

Consider a parcel of volume $v$ below the diaphragm containing $n$ moles of helium. It is convenient to imagine it bounded by two fictitious free thin massless pistons. During slow perturbations the parcel undergoes an adiabatic process. The pressure and the temperature for the parcel are actually the pressure and the temperature for entire lower chamber $p_2$ and $T_2$ . The energy conservation for the parcel is

0 = p_2\mathrm{d}v + \mathrm{d}\left(\frac{3}{2}n_vRT_2\right) = \frac{5}{2}p_2\mathrm{d}v + \frac{3}{2}v\mathrm{d}p_2.

This gives

v^5p_2^3 = \mathrm{const}, \qquad (1)

T_2^5p_2^{-2} = \mathrm{const}\,. \qquad (2)

The leak begins when the pressure below the diaphragm exceeds that in the upper chamber by $\Delta p \equiv p_0 - p = mgH/V = px$ , where

x = \frac{mgH}{pV}. \tag{\textbf{a., b., c.}}

a. We may let $v = V_2$ before that.

V^5p^3 = V_0^5(p+\Delta p)^3 = V_0^5p^3(1+x)^3

V_0 = V(1+x)^{-3/5} \tag{\textbf{a.}}

b. The energy conservation for the whole system:

0 = p_1\mathrm{d}V_1 + \mathrm{d}\left(\frac{3}{2}n_1RT_1\right) + p_2\mathrm{d}V_2 + \mathrm{d}\left(\frac{3}{2}n_2RT_2\right) =

\frac{5}{2}\left(p_1\mathrm{d}V_1 + p_2\mathrm{d}V_2\right) + \frac{3}{2}\left(V_1\mathrm{d}p_1 + V_2\mathrm{d}p_2\right) =

\frac{5}{2}p_2\mathrm{d}(V+V_2) + \frac{3}{2}(V+V_2)\mathrm{d}p_2,

since the pressure above the diaphragm remains lower than that below by the same margin $\Delta p$ during the later process and $\mathrm{d}V_2 = \mathrm{d}(V+V_2)$ . Similarly to (1), we get

(V+V_2)^5p_2^3 = \mathrm{const}\,.

The pressure $p_2'$ in the lower chamber when the piston touches the diaphragm is found from the equation

V^5p_2^{\prime 3} = (V+V_0)^5p_0^3 = V^5\left(1+\frac{1}{(1+x)^{3/5}}\right)^5p^3(1+x)^3,

p_2' = p\left(1+(1+x)^{3/5}\right)^{5/3}. \qquad (3)

The pressure in the upper chamber at this moment is

p_1' = p_2' - \Delta p = p\left(\left(1+(1+x)^{3/5}\right)^{5/3} - x\right).

The temperature in the upper chamber is found from the equations of state $pV = nRT$ and $p'V = (2n)RT'$

T_1' = \frac{T}{2}\left(\left(1+(1+x)^{3/5}\right)^{5/3} - x\right). \tag{\textbf{b.}}

The temperature and the pressure in the lower chamber are related by (2). Substituting (3) we get

T_2' = T\left(\frac{p_2'}{p}\right)^{2/5} = T\left(1+(1+x)^{3/5}\right)^{2/3}. \tag{\textbf{c.}}

Preliminary grading scheme

a1	It’s stated (or written as a formula) that the process is adiabatic	0.5
a2	Relation between $V$ and $p$ is found in adiabatic process	1.0
a3	Condition on when the diaphragm leaks	0.5
a4	Answer for $V_0$	1.0
b1	Energy conservation for the whole system in differential form. If conservation law is written only for one half but heat transfer is taken into account: 0.5 pts.	1.0
b2	Internal energy for a mono-atomic gas	0.5
b3	Usage of $V_1 = \mathrm{const}$	0.2
b3	Usage of $p_2 - p_1 = \mathrm{const}$	0.3
b4	Relation between $V_2$ and $p_2$	1.0
b5	Equation to find $p_2'$ (or $T_2'$ ) before the end	1.0
b6	Usage of $n' = 2n$	0.5
b7	Answer for $T_1'$	0.5
c1	Relation between $T_1'$ and $T_2'$	1.0
c2	Answer for $T_2'$	1.0

Arithmetic or typo errors gives half of point (rounded up to 0.1) for the item and is not considered as a mistake afterwards.

Оригинал в Архива: EuPhO_2021-theory.pdf · официални решения: EuPhO_2021_Solutions-2.pdf