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36 votes
36 votes
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

User Ben Ashton
by
3.2k points

2 Answers

22 votes
22 votes

Answer:

0.49 liters

Step-by-step explanation:

PLATO answer

User Mehulmpt
by
3.1k points
15 votes
15 votes

Answer:

0.49 L

Step-by-step 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

User Ilya Etingof
by
2.8k points