A balloon was filled to a volume of 2.50 LL when the temperature was 30.0∘C30.0∘C. What would the volume become if the temperature dropped to 11.0∘C11.0∘C.

The volume would become 2.34 L if the temperature dropped to 11.0 °C

Explanation:

From the question, A balloon was filled to a volume of 2.50 L when the temperature was 30.0 °C.

To determine what the volume will become if the temperature dropped to 11.0 °C, we will use one of the Gas laws that relates Volume and Temperature. The Gas law that relates Volume and Temperature is the Charlse' law which states that, "the Volume of a fixed mass of gas is directly proportional to its Temperature ( in Kelvin) provided that the Pressure remains constant"

That is

V ∝ T ( at constant pressure)

Where V is the Volume

and T is the Temperature in Kelvin

Then, we can write that

V = kT

Where k is the constant of proportionality

Then,

Hence, we can write that

∴

Where is the initial volume

is the initial temperature

is the final volume

and is the final temperature

From the question,

= 2.50 L

= 30.0 °C = 30.0 + 273.15 K = 303.15 K

= ??

= 11.0 °C = 11.0 + 273.15 K = 284.15 K

Putting the values into the equation, we get

∴

Hence, the volume would become 2.34 L if the temperature dropped to 11.0 °C.

d. platinum would have the greatest temperature increase because it has the lowest specific heat.