Question

A balloon is inflated with helium to a volume of 4.6L at 23.0 oC. If you leave it in a hot car, what volume will it be at 115 oF?

(a) 4.3 L

(b) 5.0 L

(c) 9.2 L

(d) 2.3 L

Answer 1

Answer:

B

Explanation:

V1 = 4.6L

T1 = 23°C = (23 + 273.15)K = 296.15

T2 = 115°F = (115 + 225.928)K = 340.928K

V2 = ?

To solve this question, we need to use Charles law which states that the volume of a fixed mass of mass is directly proportional to its temperature provided that the pressure remains constant.

Mathematically,

V = kT, k = V / T

V1 / T1 = V2 / T2 = V3 / T3 =.......=Vn / Tn

V1 / T1 = V2 / T2

Solve for V2

V2 = (V1 × T2) / T1

V2 = (4.6 × 340.928) / 296.15

V2 = 5.2L

The final of the gas is approximately 5.0L

Answer 2

Answer:

(b) 5.0 L

Explanation:

Hello,

In this case, by using the Charles' law which help us to understand the volume-temperature relationship as a directly proportional relationship, we must compute it in absolute temperatures:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$T_1=23.0+273.15=296.15K\\T_2=(115-32)\times \frac{5}{9}+273.15=319.26K$

We can easily compute the final volume by:

$V_2=\frac{V_1T_2}{T_1}=\frac{4.6L*319.26K}{296.15K}\\ \\V_2=5.0L$

Thereby, answer is (b) 5.0 L

Regards.