*Urgent*

A balloon has a volume of 31.8 L at a temperature of 46 degrees C. What is the new temperature

of the balloon in degrees C if the volume is changed to 49.2 liters?

Round answers to 0.1 decimals

A balloon has a volume of 31.8 L at a temperature of 46 degrees C. What is the new temperature

of the balloon in degrees C if the volume is changed to 49.2 liters?

Round answers to 0.1 decimals

Answer:The new volume will be

0.7 L

Explanation:This is an illustration of Charles’ law, some of the time called the temperature-volume law. It expresses that the volume of a gas is straightforwardly corresponding to the Kelvin temperature, while pressing factor and sum are held steady. The condition is V 1 T 1 = V 2 T 2 , where V is volume and T is temperature in Kelvins. Known V 1 = 0.5 L T 1 = 20 ∘ C + 273.15 = 293 K T 2 = 150 ∘ C + 273.15 = 423 K Obscure V 2 Arrangement Revise the condition to confine V 2 . Substitute the known qualities into the condition and address. V 2 = V 1 T 2 T 1 V 2 = ( 0.5 L × 423 K ) 293 K = 0.7 L adjusted to one huge figure