A man heats a balloon in the oven. If the balloon initially has a volume of 0.3 liters and a temperature of 25 0C, what will the volume of the balloon be after he heats it to a temperature of 275 0C?

0.55L

Explanation:

Given that,

Original volume of balloon (V1) = 0.3L

Original temperature of balloon (T1) = 25.0°C

[Convert 25.0°C to Kelvin by adding 273

25.0°C + 273 = 298K]

New volume of balloon (V2) = ?

New temperature of balloon (T2# = 275.0°C

[Convert 25.0°C to Kelvin by adding 273

275.0°C + 273 = 548K]

Since volume and temperature are given while pressure is held constant, apply the formula for Charle's law

V1/T1 = V2/T2

0.3L/298K = V2/548K

To get the value of V2, cross multiply

0.3L x 548K = 298K x V2

164.4L•K = 298K•V2

Divide both sides by 298K

164.4L•K /298K = 298K•V2/298K

0.55L = V2

Thus, the new volume of the balloon would be 0.55 litres

explanation:

i think option no d is the correct answer