*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

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

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