A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature 50 °C. The gas is now heated to double its temperature. The modified pressure is __ Pa.

Gas at 3.23 kPa and 50°C is heated to double its temperature in a closed cylinder.

The initial temperature of the gas is $$T_1 = 50°C = 323$$ K, and doubling the Celsius temperature gives $$T_2 = 2 \times 50 = 100°C = 373$$ K.

At constant volume the pressure and temperature are related by $$\frac{P_1}{T_1} = \frac{P_2}{T_2}$$ so that $$P_2 = P_1 \times \frac{T_2}{T_1} = 3.23 \times \frac{373}{323} = 3.23 \times 1.1548 = 3.73$$ kPa.

Converting to pascals gives $$P_2 = 3.73 \times 1000 = 3730$$ Pa.

The correct answer is $$\boxed{3730}$$ Pa.