To determine the size of the container needed for 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:
- \( P \) = pressure (in pascals)
- \( V \) = volume (in cubic meters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (\( R = 8.314 , \text{J/(mol·K)} \))
- \( T \) = temperature (in kelvins)
At STP:
- The pressure \( P = 1 , \text{ATM} = 1.013 \times 10^5 , \text{Pa} \)
- The temperature \( T = 0° \text{C} = 273.15 , \text{K} \)
- The number of moles \( n = 10 , \text{moles} \)
Now we can rearrange the ideal gas law to solve for \( V \):
\[ V = \frac{nRT}{P} \]
Plugging in the values:
\[ V = \frac{10 , \text{moles} \times 8.314 , \text{J/(mol·K)} \times 273.15 , \text{K}}{1.013 \times 10^5 , \text{Pa}} \]
Now we calculate this step-by-step:
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Calculate \( nRT \): \[ nRT = 10 \times 8.314 \times 273.15 \] \[ nRT = 2270.2221 , \text{J} \]
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Now calculate the volume \( V \): \[ V = \frac{2270.2221}{1.013 \times 10^5} \] \[ V \approx 0.02239 , \text{m}^3 \]
To convert this to liters (since 1 m³ = 1000 L):
\[ V \approx 0.02239 \times 1000 = 22.39 , \text{L} \]
Therefore, you would need a container with a volume of approximately 22.39 liters to contain 10 moles of an ideal gas at STP.