A balloon has a volume of 2.29 liters at a pressure of 100.0 kPa. If the pressure of the balloon is decreased to 83.5 kPa, what will be the new volume of the balloon?

The new volume of the balloon is 1.9 liters

The parameters given in the question are;

pressure 1= 100 kPa

pressure2= 83.5 kPa

Volume 1= 2.29 liters

P1/V1= P2/V2

100/2.29= 83.5/V2

Cross multiply

100 V2= 83.5 × 2.29

100 V2= 191.215

V2= 191.215/100

V2= 1.91 liters

Hence the new volume is 1.91 liters

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2.74 litres

Explanation:

Given that,

Original pressure of balloon (P1) = 100.0 kPa

Original volume of balloon (V1) = 2.29 litres

New pressure of balloon (P2) = 83.5 kPa

New volume of balloon (V2) = ?

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

P1V1 = P2V2

100 kPa x 2.29 L = 83.5 kPa x V2

229 kPa•L = 83.5 kPa•V2

Divide both sides by 83.5 kPa

229 kPa•L/83.5 kPa = 83.5 kPa•V2/ 83.5 kPa

2.74 L = V2

Thus, the new volume of the balloon is 2.74 litres

8 carbon atoms forming single bonds with 18 hydrogen atoms; a triple bond exists between the third and fourth carbon atom from the right

explanation: