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4 years ago

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An insulated rigid tank is divided into two compartments of different volumes. Initially, each compartment contains the same ide

al gas at identical pressure but at different temperatures and masses. The wall separating the two compartments is removed and the two gases are allowed to mix. Assuming constant specific heats, find the simplest expression for the mixture temperature written in the form
T3 = f(m1m2, m2m3, T1, T2)T 3=f( m 2m 1, m 3m 2,T 1,T 2) where m3m 3and T3T 3are the mass and temperature of the final mixture, respectively.

1 answer:

storchak [24]4 years ago

8 0

Answer:

Explanation:

Given that

Mass of 1 = $m_1$

Mass of 2 = $m_2$

Temperature in 1 = $T_1$

Temperature in 2 = $T_2$

Pressure remains i the group apartment

The closed system and energy balance is

$E_{in}-E_{out}=\Delta E_{system}$

The kinetic energy and potential energy are negligible

since it is insulated tank ,there wont be eat transfer from the system

And there is no work involved

$\Delta U = 0$

Let the final temperature be final temperature

$m_1+c_v(T_3-T_1)+m_2c_v(T_3+T_2)=0\\\\m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)---(i)$

Using mass balance

$m_3+m_2+m_1$

from eqn i

$m_1+c_v(T_3-T_1)=m_2c_v(T_2-T_3)m_1T_3-m_1T_1=m_2T_2-m_2T_3\\\\m_1T_3+m_2T_3=m_2T_2+m_1T_1\\\\(m_1+m_2)T_3=m_1T_1+m_2T_2\\\\(m_1)T_3=m_1T_1+m_2T_2\\\\T_3=\frac{m_1T_1_m_2T_2}{m_3}$

Therefore the final temperature can be express as

$\large \boxed {T_3=\frac{m_1}{m_3} T_1+\frac{m_2}{m_3}T_2 }$

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