Question

An insulated rigid tank is divided into two compartments of different volumes. Initially, each compartment contains the same ideal 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 $$T_{3}=f\left(\frac{m_{1}}{m_{3}}, \frac{m_{2}}{m_{3}}, T_{1}, T_{2}\right)$$ where \(m_{3}\) and \(T_{3}\) are the mass and temperature of the final mixture, respectively.

Short answer

Answer: The expression for the final temperature, \(T_3\), is: $$T_3 = \frac{T_1 \frac{m_1}{m_3} + T_2 \frac{m_2}{m_3}}{\frac{m_1}{m_3}+\frac{m_2}{m_3}}$$ Here, \(T_1\) and \(T_2\) are the initial temperatures of the two compartments, \(m_1\) and \(m_2\) are their respective masses, and \(m_3\) is the total mass (sum of \(m_1\) and \(m_2\)).

Step-by-step solution

Initial Internal Energy

Calculate the initial internal energy of the gas in both compartments. For an ideal gas, the internal energy (\(U\)) is given by the formula: $$U = m c_v T$$ where \(m\) is the mass of the gas, \(c_v\) is the constant volume specific heat, and \(T\) is the temperature. For each compartment, the initial internal energy is: $$U_1 = m_1 c_v T_1$$ $$U_2 = m_2 c_v T_2$$

Apply Energy Conservation

As the gases mix, no energy is transferred to or from the system due to the insulation, so the internal energy before and after mixing should be equal. Therefore, $$U_1 + U_2 = U_3$$

Final Internal Energy

Calculate the final internal energy. The mass, \(m_3\), of the mixture is the sum of the two individual masses, so $$m_3 = m_1 + m_2$$ The internal energy of the mixture can be expressed as $$U_3 = m_3 c_v T_3$$

Solve for Final Temperature

Using the relationships in steps 1, 2, and 3, we can solve for the final temperature \(T_3\). First, let's rewrite the energy conservation equation in terms of mass and temperature:$$m_1 c_v T_1 + m_2 c_v T_2 = (m_1 + m_2) c_v T_3$$The specific heat values can be canceled out:$$m_1 T_1 + m_2 T_2 = (m_1 + m_2) T_3$$Now, divide both sides by \(m_3\) and factor out \(T_3\) to get:$$T_3 \frac{m_1}{m_3} + T_3 \frac{m_2}{m_3} = T_1 \frac{m_1}{m_3} + T_2 \frac{m_2}{m_3}$$Finally, we find the expression for the final temperature, \(T_3\):$$T_3 = \frac{T_1 \frac{m_1}{m_3} + T_2 \frac{m_2}{m_3}}{\frac{m_1}{m_3}+\frac{m_2}{m_3}}$$