Solution:

Given:

- Initial temperature, \( T_1 = 27^\circ \text{C} = 300 \, \text{K} \)
- We want to expel half the mass of air, meaning the final mass, \( m_2 = \frac{m_1}{2} \).

Using the ideal gas law, \( PV = nRT \), and since pressure and volume are constant, \( \frac{m}{T} = \text{constant} \).

Thus,

\[
\frac{m_1}{T_1} = \frac{m_2}{T_2}
\]

Substitute \( m_2 = \frac{m_1}{2} \):

\[
\frac{m_1}{300} = \frac{\frac{m_1}{2}}{T_2}
\]

Solving for \( T_2 \):

\[
T_2 = 2 \times 300 = 600 \, \text{K}
\]

Converting back to Celsius:

\[
T_2 = 600 - 273 = 327^\circ \text{C}
\]

Answer: \( T_2 = 327^\circ \text{C} \)