A husband buys a helium-filled anniversary balloon for his wife. The balloon has a volume of 4.4 L in the warm store at 74 degree F. When he takes it outside, where the temperature is 55 degree F, he finds it has shrunk. By how much has the volume decreased?

This is what I did:

74-32=45(5/9)=23+273=296K
55-32=45(5/9)=12.8+273=286K

4.4/296 = V2/286
286(4.4)/296 = 4.25L

But I'm getting the wrong answer. What am I doing wrong? Help please.

To convert Fahrenheit to Rankine, add 460.

74 F = 534 R = 297 K

55 F = 515 R = 286 K

If pressure stayed the same, V/T remains constant and

V2 = (286/297)4.4 = 4.24 L

Actually, a falling volume decreases the pressure in an already-stretched balloon. One would need more data on balloon tension and elasticity to take this into account.

This may help.

http://www.digipac.ca/chemical/gaslaws/TchrNotesPVB.htm

Your calculations are correct, but there is a small error in the final calculation. Let's go through the steps again to find the correct answer.

First, convert the temperatures from Fahrenheit to Kelvin:
74°F - 32 = 42°F
42°F × 5/9 = 23.33°C
23.33°C + 273.15 = 296.48K

55°F - 32 = 23°F
23°F × 5/9 = -5°C
-5°C + 273.15 = 268.15K

Now, let's use the ideal gas law equation to find the new volume (V2):
P1V1/T1 = P2V2/T2

Since the pressure (P1) and pressure (P2) remain constant, we can simplify the equation:
V1/T1 = V2/T2

Plug in the values:
4.4L / 296.48K = V2 / 268.15K

Now, cross multiply and solve for V2:
V2 = (4.4L × 268.15K) / 296.48K
V2 = 3.972L (rounded to three decimal places)

Therefore, the volume of the balloon has decreased by approximately 0.428L (4.4L - 3.972L) when taken outside.

To solve this problem, you need to apply Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is constant. The equation for Charles's Law is as follows:

V1/T1 = V2/T2

where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.

Let's first convert the temperatures to Kelvin (K) since the Kelvin scale is used in gas laws.

Given:
T1 = 74°F = 23.3°C = 296.3K
T2 = 55°F = 12.8°C = 285.9K

Now, we can substitute the values into the equation:

V1/T1 = V2/T2

V1/296.3 = V2/285.9

Now, let's solve for V2:

V2 = (V1 * 285.9) / 296.3

Plug in the value of V1, which is 4.4L:

V2 = (4.4 * 285.9) / 296.3

Calculating this gives us the final volume:

V2 ≈ 4.237 L (rounded to three decimal places)

So, the volume of the balloon has decreased by approximately 0.163 L (4.4 L - 4.237 L = 0.163 L), not 0.25 L as you calculated.

Therefore, the correct answer is that the volume of the balloon has decreased by approximately 0.163 L.