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APhO 2006, Теория — Задача 3B. Part B

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Всички материали от APhO 2006

7th Asian Physics Olympiad — Theoretical Question 3 · 2,5 т.

Условие

B. [2.5 points] Light mobile piston separates the vessel into two parts. The vessel is isolated from the environment. One part of the vessel contains $m_1 = 3g$ of hydrogen at the temperature of $T_{10} = 300\ \mathrm{K}$ , and the other part contains $m_2 = 16\ g$ of oxygen at the temperature of $T_{20} = 400\ \mathrm{K}$ . Molar masses of hydrogen and oxygen are $\mu_1 = 2\ g/mole$ and $\mu_2 = 32\ g/mole$ respectively, and $R = 8.31 J/(K\cdot mole)$ . The piston weakly conducts heat between oxygen and hydrogen, and eventually the temperature in the system equilibrates. All the processes are quasi stationary.

B1) What is the final temperature of the system $T$ ?

B2) What is the ratio between final pressure $P_f$ and initial pressure $P_i$ ?

B3) What is the total amount of heat $Q$ , transferred from oxygen to hydrogen?

Отговори

Решение

The total work done by the gases is zero. Thus at any moment the total internal energy equals the original value:

\frac{m_1}{\mu_1}C_V T_1 + \frac{m_2}{\mu_2}C_V T_2 = \frac{m_1}{\mu_1}C_V T_{10} + \frac{m_2}{\mu_2}C_V T_{20}, \qquad (13)

where $\mu_1 = 2\ g/mole$ and $\mu_2 = 32\ g/mole$ are molar masses of hydrogen and oxygen, and $C_V=5R/2$ is the molar heat capacity of diatomic gas. The final temperature of the system is

(B1)

T = \frac{\frac{m_1}{\mu_1}T_{10} + \frac{m_2}{\mu_2}T_{20}}{\frac{m_1}{\mu_1} + \frac{m_2}{\mu_2}} = 325\ K. \qquad (14)

The temperature of oxygen decreases, and the amount of heat Q is transferred to hydrogen by heat conduction. The piston will move in the direction of the oxygen, thus the hydrogen does a positive work $A>0$ , and the change of the internal energy of oxygen is $\Delta U = A - Q$ . On the other hand,

\Delta U = \frac{m_2}{\mu_2}\frac{5}{2}R(T - T_{20}) = -779\ J. \qquad (15)

To find $A$ , let us prove that the pressure $P$ doesn’t change. Differentiating the equations of the state for each gas, we get

\Delta T_1 = \frac{\mu_1}{m_1 R}\left(P\Delta V + V_1\Delta P\right), \quad \Delta T_2 = \frac{\mu_2}{m_2 R}\left(-P\Delta V + V_2\Delta P\right). \qquad (16)

where $V_i$ are the gas volumes, and $\Delta V = \Delta V_1 = -\Delta V_2$ is the change of the volume of the hydrogen. Differentiating (1), we get

\frac{m_1}{\mu_1}\Delta T_1 + \frac{m_2}{\mu_2}\Delta T_2 = 0. \qquad (17)

Substituting (2) into (3), we obtain $(V_1+V_2)\cdot\Delta P = 0$ , thus

(B2)

P_f/P_i = 1. \qquad (18)

Then the work done by the hydrogen is

A = P\cdot\Delta V = -\frac{m_2}{\mu_2}R\cdot\Delta T_2 = \frac{m_2}{\mu_2}R(T_{20} - T) = 312\ J. \qquad (19)

The total amount of heat transferred to hydrogen is

(B3)

Q = A - \Delta U = 1091 J. \qquad (20)

Mark Distribution

No	Total Pt	Partial Pt	Contents
B	2.5	0.5	Numerical value of T
		1.0	Constant pressure
		0.3	Numerical value of $\Delta U$
		0.3	Numerical value of $A$
		0.4	Numerical value of $Q$

Оригинал в Архива: 3.pdf · официални решения: 3_sol.pdf