A high altitude balloon contains 6.81 g of helium in `1.16xx10^(4)" L at"-23^(@)C.` Assuming ideal gas behaviour, how many grams of helium would have to be added to increase the pressure to `4.0xx10^(-3)` atm?

Options

A. 1.27 g
B. 1.58 g
C. 2.68 g
D. 2.23 g

Explanation

To solve the problem step by step, we will use the ideal gas law equation, which is given by:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature (in Kelvin)

### Step 1: Convert Temperature to Kelvin
The temperature is given as -23°C. To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

So,

\[ T = -23 + 273.15 = 250.15 \, K \]

### Step 2: Use the Ideal Gas Law to Find Initial Moles of Helium
We will use the ideal gas law to find the number of moles of helium at the initial conditions.

Given:
- \( P = 4.0 \times 10^{-3} \, \text{atm} \)
- \( V = 1.16 \times 10^{4} \, \text{L} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 250.15 \, K \)

Substituting these values into the ideal gas equation:

\[ n = \frac{PV}{RT} \]

\[ n = \frac{(4.0 \times 10^{-3} \, \text{atm})(1.16 \times 10^{4} \, \text{L})}{(0.0821 \, \text{L·atm/(K·mol)})(250.15 \, K)} \]

Calculating the numerator:

\[ 4.0 \times 10^{-3} \times 1.16 \times 10^{4} = 46.4 \]

Calculating the denominator:

\[ 0.0821 \times 250.15 = 20.525 \]

Now, substituting these values:

\[ n = \frac{46.4}{20.525} \approx 2.26 \, \text{moles} \]

### Step 3: Calculate the Mass of Helium Required
The molar mass of helium (He) is approximately 4 g/mol. To find the mass of helium required, we use the formula:

\[ W = n \times \text{molar mass} \]

So,

\[ W = 2.26 \, \text{moles} \times 4 \, \text{g/mol} = 9.04 \, \text{g} \]

### Step 4: Determine the Mass of Helium to be Added
The balloon already contains 6.81 g of helium. To find out how much more helium needs to be added, we subtract the mass already present from the total mass required.

\[ \text{Mass to be added} = W - \text{mass already present} \]

\[ \text{Mass to be added} = 9.04 \, \text{g} - 6.81 \, \text{g} = 2.23 \, \text{g} \]

### Final Answer
To increase the pressure to \( 4.0 \times 10^{-3} \) atm, **2.23 grams of helium** would have to be added.

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