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Answer :
We are given that a balloon has an initial volume
[tex]$$V_1 = 43 \text{ L}$$[/tex]
at an initial temperature
[tex]$$T_1 = 25^\circ\text{C}.$$[/tex]
The balloon is then cooled to a new temperature
[tex]$$T_2 = -8^\circ\text{C}.$$[/tex]
Since the pressure remains constant, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (measured in Kelvin). Charles's Law is expressed as:
[tex]$$
\frac{V_1}{T_1} = \frac{V_2}{T_2}.
$$[/tex]
### Step 1. Convert the Temperatures to Kelvin
To convert from Celsius to Kelvin, we use the formula:
[tex]$$
T(\text{K}) = T(^\circ\text{C}) + 273.15.
$$[/tex]
Thus,
[tex]$$
T_1 = 25 + 273.15 = 298.15 \text{ K},
$$[/tex]
[tex]$$
T_2 = -8 + 273.15 = 265.15 \text{ K}.
$$[/tex]
### Step 2. Solve for the New Volume [tex]$V_2$[/tex]
Rearrange Charles's Law to solve for [tex]$V_2$[/tex]:
[tex]$$
V_2 = V_1 \times \frac{T_2}{T_1}.
$$[/tex]
Substitute the values:
[tex]$$
V_2 = 43 \times \frac{265.15}{298.15}.
$$[/tex]
### Step 3. Calculate the Temperature Ratio and [tex]$V_2$[/tex]
First, calculate the ratio:
[tex]$$
\frac{T_2}{T_1} = \frac{265.15}{298.15} \approx 0.8893.
$$[/tex]
Now, compute [tex]$V_2$[/tex]:
[tex]$$
V_2 \approx 43 \times 0.8893 \approx 38.24 \text{ L}.
$$[/tex]
When rounding to one decimal place, the volume is approximately:
[tex]$$
V_2 \approx 38.2 \text{ L}.
$$[/tex]
### Final Answer
The volume of the balloon at [tex]$-8^\circ\text{C}$[/tex] is approximately
[tex]$$\boxed{38.2 \text{ L}}.$$[/tex]
[tex]$$V_1 = 43 \text{ L}$$[/tex]
at an initial temperature
[tex]$$T_1 = 25^\circ\text{C}.$$[/tex]
The balloon is then cooled to a new temperature
[tex]$$T_2 = -8^\circ\text{C}.$$[/tex]
Since the pressure remains constant, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (measured in Kelvin). Charles's Law is expressed as:
[tex]$$
\frac{V_1}{T_1} = \frac{V_2}{T_2}.
$$[/tex]
### Step 1. Convert the Temperatures to Kelvin
To convert from Celsius to Kelvin, we use the formula:
[tex]$$
T(\text{K}) = T(^\circ\text{C}) + 273.15.
$$[/tex]
Thus,
[tex]$$
T_1 = 25 + 273.15 = 298.15 \text{ K},
$$[/tex]
[tex]$$
T_2 = -8 + 273.15 = 265.15 \text{ K}.
$$[/tex]
### Step 2. Solve for the New Volume [tex]$V_2$[/tex]
Rearrange Charles's Law to solve for [tex]$V_2$[/tex]:
[tex]$$
V_2 = V_1 \times \frac{T_2}{T_1}.
$$[/tex]
Substitute the values:
[tex]$$
V_2 = 43 \times \frac{265.15}{298.15}.
$$[/tex]
### Step 3. Calculate the Temperature Ratio and [tex]$V_2$[/tex]
First, calculate the ratio:
[tex]$$
\frac{T_2}{T_1} = \frac{265.15}{298.15} \approx 0.8893.
$$[/tex]
Now, compute [tex]$V_2$[/tex]:
[tex]$$
V_2 \approx 43 \times 0.8893 \approx 38.24 \text{ L}.
$$[/tex]
When rounding to one decimal place, the volume is approximately:
[tex]$$
V_2 \approx 38.2 \text{ L}.
$$[/tex]
### Final Answer
The volume of the balloon at [tex]$-8^\circ\text{C}$[/tex] is approximately
[tex]$$\boxed{38.2 \text{ L}}.$$[/tex]
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