The condition called standard temperature and pressure (STP) for a gas is defined as temperature of `0^@ C = 273.15 K` and a pressure of `1 atm = 1.013 xx 10^5 Pa`. If you want to keep a mole of an ideal gas in your room at (STP), how big a container do you need ?

Explanation

To determine the size of the container needed to hold one mole of an ideal gas at standard temperature and pressure (STP), we can use the Ideal Gas Law, which is expressed as:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in Pascals)
- \( V \) = volume (in cubic meters)
- \( n \) = number of moles of gas
- \( R \) = universal gas constant (approximately \( 8.314 \, \text{J/(mol K)} \))
- \( T \) = temperature (in Kelvin)

### Step-by-Step Solution:

1. **Identify the known values:**
- At STP, the temperature \( T = 273.15 \, \text{K} \)
- The pressure \( P = 1.013 \times 10^5 \, \text{Pa} \)
- The number of moles \( n = 1 \, \text{mol} \)
- The universal gas constant \( R = 8.314 \, \text{J/(mol K)} \)

2. **Rearrange the Ideal Gas Law to solve for volume \( V \):**
\[ V = \frac{nRT}{P} \]

3. **Substitute the known values into the equation:**
\[ V = \frac{(1 \, \text{mol}) \times (8.314 \, \text{J/(mol K)}) \times (273.15 \, \text{K})}{1.013 \times 10^5 \, \text{Pa}} \]

4. **Calculate the numerator:**
\[ 1 \times 8.314 \times 273.15 = 2270.24 \, \text{J} \]

5. **Now calculate the volume \( V \):**
\[ V = \frac{2270.24 \, \text{J}}{1.013 \times 10^5 \, \text{Pa}} \approx 0.0224 \, \text{m}^3 \]

6. **Convert the volume to a more convenient unit if necessary:**
\[ V \approx 2.24 \times 10^{-2} \, \text{m}^3 \]

### Final Answer:
To hold one mole of an ideal gas at standard temperature and pressure, you would need a container with a volume of approximately \( 2.24 \times 10^{-2} \, \text{m}^3 \) or \( 22.4 \, \text{liters} \).