To determine the size of the container needed to hold 10 moles of an ideal gas at standard temperature and pressure (STP), we can use the Ideal Gas Law, which is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) is the pressure (in Pascals)
- \( V \) is the volume (in cubic meters)
- \( n \) is the number of moles of gas
- \( R \) is the ideal gas constant \( (8.314 , \text{J/(mol·K)}) \)
- \( T \) is the temperature (in Kelvin)
At STP:
- \( P = 1 , \text{atm} = 1.013 \times 10^5 , \text{Pa} \)
- \( T = 273 , \text{K} \)
- \( n = 10 , \text{moles} \)
- \( R = 8.314 , \text{J/(mol·K)} \)
We can rearrange the Ideal Gas Law to solve for \( V \):
\[ V = \frac{nRT}{P} \]
Substituting the known values into the equation:
\[ V = \frac{(10 , \text{mol}) (8.314 , \text{J/(mol·K)}) (273 , \text{K})}{1.013 \times 10^5 , \text{Pa}} \]
Calculating the numerator:
\[ (10) (8.314) (273) = 22700.22 , \text{J} \]
Now, calculating the volume \( V \):
\[ V = \frac{22700.22 , \text{J}}{1.013 \times 10^5 , \text{Pa}} \approx 0.223 , \text{m}^3 \]
Thus, the volume required to contain 10 moles of an ideal gas at STP is approximately:
\[ \boxed{0.223 , \text{m}^3} \]
or about 223 liters, since \( 1 , \text{m}^3 = 1000 , \text{liters} \).