A partially inflated balloon contains `500 m^3` of helium at `27^@ C` and `1 atm` pressure. What is the volume of the helium at an altitude of `18000 ft`, where the pressure is `0.5 atm` and the temperature is `-3^@ C` ?

Explanation

To solve the problem of finding the volume of helium in a balloon at a higher altitude with different pressure and temperature, we can use the Ideal Gas Law, which states that:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Where:
- \(P_1\) = initial pressure
- \(V_1\) = initial volume
- \(T_1\) = initial temperature (in Kelvin)
- \(P_2\) = final pressure
- \(V_2\) = final volume
- \(T_2\) = final temperature (in Kelvin)

### Step 1: Convert temperatures from Celsius to Kelvin
To convert Celsius to Kelvin, use the formula:
\[ T(K) = T(°C) + 273.15 \]
- For \(T_1 = 27°C\):
\[ T_1 = 27 + 273.15 = 300.15 \, K \]
- For \(T_2 = -3°C\):
\[ T_2 = -3 + 273.15 = 270.15 \, K \]

### Step 2: Identify the values
From the problem, we have:
- \(V_1 = 500 \, m^3\)
- \(P_1 = 1 \, atm\)
- \(P_2 = 0.5 \, atm\)
- \(T_1 = 300.15 \, K\)
- \(T_2 = 270.15 \, K\)

### Step 3: Rearrange the Ideal Gas Law to solve for \(V_2\)
We can rearrange the equation to find \(V_2\):
\[ V_2 = V_1 \cdot \frac{P_1}{P_2} \cdot \frac{T_2}{T_1} \]

### Step 4: Substitute the known values into the equation
\[ V_2 = 500 \, m^3 \cdot \frac{1 \, atm}{0.5 \, atm} \cdot \frac{270.15 \, K}{300.15 \, K} \]

### Step 5: Calculate \(V_2\)
Calculating the fractions:
\[ V_2 = 500 \cdot 2 \cdot \frac{270.15}{300.15} \]
Calculating \( \frac{270.15}{300.15} \):
\[ \frac{270.15}{300.15} \approx 0.900 \]
Now substituting back:
\[ V_2 = 500 \cdot 2 \cdot 0.900 \approx 900 \, m^3 \]

### Final Answer
The volume of the helium at an altitude of 18,000 ft is approximately \(900 \, m^3\).