Question

Dwayne filled a small balloon with air at 298.5 K. He put the balloon into a bucket of water, and the water level in the bucket
increased by 0.54 liter.
If Dwayne puts the balloon into a bucket of ice water at 273.15 K and waits for the air inside the balloon to come to the same
temperature, what will the volume of the balloon be? Assume the pressure inside the balloon doesn't change.
Type the correct answer in the box. Express your answer to the correct number of significant figures e.
The volume of the balloon at 273.15 K is
I liters.

CORRECT ANSWER:
.49

Answer 1

Answer:

0.49 L

Explanation:

T1 = 298.5 K.

V1= 0.54 L

T2= 273.15 K

V2= ?

From Charles law;

V1/T1 = V2/T2

V1T2= V2T1

V2= V1T2/T1

V2= 0.54 × 273.15/298.5

V2= 0.49 L

Answer 2

Answer:

0.49 liters

Explanation:

PLATO answer