Question

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.

Answer 1

Answer:

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,

$\frac{V}{T} = k$

Hence, we can write that

$\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} } = \frac{V_{3} }{T_{3} } ...$

∴ $\frac{V_{1} }{T_{1} } = \frac{V_{2} }{T_{2} }$

Where ${V_{1} }$ is the initial volume

${T_{1} }$ is the initial temperature

${V_{2} }$ is the final volume

and ${T_{2} }$ is the final temperature

From the question,

${V_{1} }$ = 2.50 L

${T_{1} }$ = 30.0 °C = 30.0 + 273.15 K = 303.15 K

${V_{2} }$ = ??

${T_{2} }$ = 11.0 °C = 11.0 + 273.15 K = 284.15 K

Putting the values into the equation, we get

$\frac{2.50}{303.15} = \frac{V_{2} }{284.15}$

∴ $V_{2} = \frac{2.50 \times 284.15}{303.15}$

$V_{2} = 2.34 L$

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