A student increases the temperature of a 300 cm^3 balloon from 30c to 60c. what will the new volume of the ballon be

From Charles law the volume of a fixed mass of gas is directly proportional to the absolute temperature when pressure is kept constant.

Volume α temperature, mathematically; k = V/T

Therefore, comparing two gases V1/T1 = V2/T2

V1 = 300 cm³ T1 = 30°C (30 + 273 = 303K) V2 = ? T2 = 60°C (60 + 273 = 333K)

Thus, 300/303 = V2/333

V2 = (300 * 333) / 303

= 329.703 cm³

Hence the new volume will be 329.703 cm³

This problem uses the ideal gas law, PV = nRT. T for temperature must be in Kelvin, so we add 273.15 to the Celsius temperatures. We are then changing T from 303.15 K to 333.15K. The ratio of the temperatures 333.15/303.15 ~ = 1.1. So holding pressure constant, the new volume of the balloon will be approximately 10% larger, or 330cm^3.