The increase in volume of air, when temperature of \(600 \mathrm{ml}\) of it, is increased from \(27^{\circ} \mathrm{C}\) to \(47^{\circ} \mathrm{C}\) under constant pressure, is : (1) \(20 \mathrm{~mL}\) (2) \(80 \mathrm{~mL}\) (3) \(40 \mathrm{~mL}\) (4) \(500 \mathrm{~mL}\)

When studying gases, one of the most important principles to understand is the gas laws. These laws describe the behavior of gases in terms of volume, temperature, and pressure. For this exercise, we're focusing on Charles's Law. Charles's Law states that at constant pressure, the volume of a gas is directly proportional to its temperature, which we can express with the formula: \[\frac{V_1}{T_1} = \frac{V_2}{T_2}\]. This means that as the temperature of a gas increases, its volume also increases, provided the pressure stays the same. The inverse is also true: if the temperature decreases, so does the volume. To correctly apply this law, it’s crucial to first convert temperatures from degrees Celsius to Kelvin, since the absolute temperature scale is required.

Temperature conversion is a vital step in solving gas law problems. Gases react based on their absolute temperature, measured in Kelvin. To convert Celsius to Kelvin, simply add 273.15 to the Celsius temperature. For example, the initial temperature given in the problem is \(27^{\text{C}}\). Converting this to Kelvin involves adding 273.15: \[27 + 273.15 = 300.15 \text{ K}\]. Similarly, the final temperature of \(47^{\text{C}}\) converts as: \[47 + 273.15 = 320.15 \text{ K}\]. This conversion ensures that all calculations follow the correct scientific standard, leading to accurate results.

Once temperatures are converted, you can apply Charles's Law to calculate new volumes. In our problem, we know the initial volume (\(600 \text{ mL}\)) and both the initial (\(300.15 \text{ K}\)) and final (\(320.15 \text{ K}\)) temperatures. Using the formula \[V_2 = V_1 \times \frac{T_2}{T_1}\], we substitute the values: \[V_2 = 600 \times \frac{320.15}{300.15}\]. Performing the calculation gives us a final volume of approximately \(640 \text{ mL}\). The increase in volume, \(\bigtriangleup V\), is then determined by subtracting the initial volume from the final volume: \[640 - 600 = 40 \text{ mL}\]. This properly solves the problem and confirms the correct answer as \(40 \text{ mL}\).