To solve this problem, we can use the combined gas law in terms of pressure ($P$) and temperature ($T$). For a gas with a fixed volume, the relationship is given as:

$$ \frac{P_1}{T_1} = \frac{P_2}{T_2} $$

where:

$ P_1 $ is the initial pressure,

$ T_1 $ is the initial temperature in Kelvin,

$ P_2 $ is the final pressure,

$ T_2 $ is the final temperature in Kelvin.

Given that the initial temperature $ T_1 = 27°C $, we need to convert it to Kelvin:

$$ T_1 = 27°C + 273.15 = 300.15 \, \text{K} $$

The final pressure $ P_2 $ is twice the initial pressure ($ P_2 = 2P_1 $). Substituting these values into the gas law equation gives:

$$ \frac{P_1}{300.15} = \frac{2P_1}{T_2} $$

Solving for $ T_2 $:

Cancel $ P_1 $ from both sides:

$$ \frac{1}{300.15} = \frac{2}{T_2} $$

Rearrange to solve for $ T_2 $:

$$ T_2 = 2 \times 300.15 = 600.3 \, \text{K} $$

Convert the final temperature back to Celsius:

$$ T_2 = 600.3 \, \text{K} - 273.15 = 327.15°C $$

Therefore, to double the pressure of the gas, the temperature should be raised to approximately 327°C.