Question
The standard temperature and pressure for a gas is define to be a temperature of 0 degree celsius (273k) and pressure of 1 ATM (1.013×10⁵ Pa). If you want to keep 10 moles of an ideal gas in your room at STP, how big is a container do you need
Answers
GPT-4o mini
To determine the size of a container required 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 formula:
\[
PV = nRT
\]
Where:
- \( P \) = pressure (in pascals)
- \( V \) = volume (in cubic meters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (approximately \( 8.314 \, \text{J/(mol K)} \))
- \( T \) = temperature (in kelvin)
Given:
- \( P = 1.013 \times 10^5 \, \text{Pa} \) (1 ATM)
- \( n = 10 \, \text{moles} \)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 273 \, \text{K} \)
Rearranging the ideal gas law to solve for volume \( V \):
\[
V = \frac{nRT}{P}
\]
Now substituting in the values:
\[
V = \frac{10 \, \text{moles} \times 8.314 \, \text{J/(mol K)} \times 273 \, \text{K}}{1.013 \times 10^5 \, \text{Pa}}
\]
Calculating the numerator:
\[
10 \times 8.314 \times 273 \approx 22700.56 \, \text{J}
\]
Now substituting the values back into the equation:
\[
V \approx \frac{22700.56 \, \text{J}}{1.013 \times 10^5 \, \text{Pa}}
\]
Calculating this:
\[
V \approx 0.224 \, \text{m}^3
\]
Thus, to hold 10 moles of an ideal gas at standard temperature and pressure (STP), you would need a container that has a volume of approximately **0.224 cubic meters**, which is equivalent to about 224 liters.
\[
PV = nRT
\]
Where:
- \( P \) = pressure (in pascals)
- \( V \) = volume (in cubic meters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (approximately \( 8.314 \, \text{J/(mol K)} \))
- \( T \) = temperature (in kelvin)
Given:
- \( P = 1.013 \times 10^5 \, \text{Pa} \) (1 ATM)
- \( n = 10 \, \text{moles} \)
- \( R = 8.314 \, \text{J/(mol K)} \)
- \( T = 273 \, \text{K} \)
Rearranging the ideal gas law to solve for volume \( V \):
\[
V = \frac{nRT}{P}
\]
Now substituting in the values:
\[
V = \frac{10 \, \text{moles} \times 8.314 \, \text{J/(mol K)} \times 273 \, \text{K}}{1.013 \times 10^5 \, \text{Pa}}
\]
Calculating the numerator:
\[
10 \times 8.314 \times 273 \approx 22700.56 \, \text{J}
\]
Now substituting the values back into the equation:
\[
V \approx \frac{22700.56 \, \text{J}}{1.013 \times 10^5 \, \text{Pa}}
\]
Calculating this:
\[
V \approx 0.224 \, \text{m}^3
\]
Thus, to hold 10 moles of an ideal gas at standard temperature and pressure (STP), you would need a container that has a volume of approximately **0.224 cubic meters**, which is equivalent to about 224 liters.