Question

How do I calculate the amount of oxygen gas in a 25.0 L container at 25.0°C and 2.50 atm using the Ideal Gas Law?

Answer 1

To calculate the amount of oxygen gas in a container using the Ideal Gas Law, you can use the formula:

\[ PV = nRT \]

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

Here are the steps to solve for \( n \) (the amount of oxygen gas in moles):

1. **Convert the temperature from Celsius to Kelvin**:
\[ T(K) = T(°C) + 273.15 \]
\[ T = 25.0 + 273.15 = 298.15 \, K \]

2. **Substitute the values into the Ideal Gas Law**:
- \( P = 2.50 \, \text{atm} \)
- \( V = 25.0 \, \text{L} \)
- \( R = 0.0821 \, \text{L·atm/(K·mol)} \)
- \( T = 298.15 \, \text{K} \)

The equation becomes:
\[ (2.50 \, \text{atm}) \times (25.0 \, \text{L}) = n \times (0.0821 \, \text{L·atm/(K·mol)}) \times (298.15 \, \text{K}) \]

3. **Calculate \( n \)**:
\[ 62.5 \, \text{atm·L} = n \times (24.475 \, \text{L·atm/(K·mol)}) \]

(where \( 0.0821 \times 298.15 = 24.475 \))

Now, solve for \( n \):
\[ n = \frac{62.5}{24.475} \]

4. **Perform the division**:
\[ n \approx 2.55 \, \text{moles} \]

So, the amount of oxygen gas in the 25.0 L container at 25.0°C and 2.50 atm is approximately **2.55 moles**.