Question
Find the total translational kinetic energy of 0.5 L of oxygen gas held at a temperature of −3 ◦C and a pressure of 3 atm. Answer in units of J.
Answers
Answered by
GPT-4o mini
To find the total translational kinetic energy of the oxygen gas, we can use the following formula for the average translational kinetic energy per mole of gas:
\[
KE_{\text{avg}} = \frac{3}{2} RT
\]
Where:
- \( R \) is the universal gas constant, which is \( 8.314 \, \text{J/(mol·K)} \)
- \( T \) is the absolute temperature in Kelvin.
First, we need to convert the given temperature from Celsius to Kelvin:
\[
T(K) = T(°C) + 273.15 = -3 + 273.15 = 270.15 \, \text{K}
\]
Next, we calculate the number of moles of oxygen gas in 0.5 L at 3 atm using the ideal gas law:
\[
PV = nRT
\]
Where:
- \( P \) is the pressure (3 atm),
- \( V \) is the volume (0.5 L = 0.5 \, \text{L} \times 0.001 \, \text{m}^3/\text{L} = 0.0005 \, \text{m}^3\)),
- \( n \) is the number of moles,
- \( R \) is the universal gas constant in suitable units (we can use \( 0.0821 \, \text{L·atm/(mol·K)} \) for this calculation).
Rearranging the ideal gas law to solve for \( n \):
\[
n = \frac{PV}{RT}
\]
Substituting in the values:
\[
n = \frac{(3 \, \text{atm})(0.5 \, \text{L})}{(0.0821 \, \text{L·atm/(mol·K)})(270.15 \, \text{K})}
\]
Calculating the denominator:
\[
\text{denominator} = 0.0821 \times 270.15 \approx 22.196 \, \text{L·atm/(mol·K)}
\]
Now substituting back into the equation for \( n \):
\[
n = \frac{1.5 \, \text{atm·L}}{22.196 \, \text{L·atm/(mol·K)}} \approx 0.0674 \, \text{mol}
\]
Now, we can calculate the total translational kinetic energy of the gas using:
\[
KE_{\text{total}} = n \cdot KE_{\text{avg}} = n \cdot \frac{3}{2} RT
\]
Substituting the values into the kinetic energy equation:
\[
KE_{\text{total}} = 0.0674 \, \text{mol} \cdot \frac{3}{2} (8.314 \, \text{J/(mol·K)}) (270.15 \, \text{K})
\]
Calculating the average kinetic energy:
\[
KE_{\text{avg}} = \frac{3}{2} \cdot 8.314 \cdot 270.15 \approx \frac{3}{2} \cdot 2242.946 \approx 3364.42 \, \text{J/mol}
\]
Now substituting back to find the total kinetic energy:
\[
KE_{\text{total}} = 0.0674 \cdot 3364.42 \approx 226.43 \, \text{J}
\]
Thus, the total translational kinetic energy of 0.5 L of oxygen gas at -3 °C and 3 atm is approximately:
\[
\boxed{226.43 \, \text{J}}
\]
\[
KE_{\text{avg}} = \frac{3}{2} RT
\]
Where:
- \( R \) is the universal gas constant, which is \( 8.314 \, \text{J/(mol·K)} \)
- \( T \) is the absolute temperature in Kelvin.
First, we need to convert the given temperature from Celsius to Kelvin:
\[
T(K) = T(°C) + 273.15 = -3 + 273.15 = 270.15 \, \text{K}
\]
Next, we calculate the number of moles of oxygen gas in 0.5 L at 3 atm using the ideal gas law:
\[
PV = nRT
\]
Where:
- \( P \) is the pressure (3 atm),
- \( V \) is the volume (0.5 L = 0.5 \, \text{L} \times 0.001 \, \text{m}^3/\text{L} = 0.0005 \, \text{m}^3\)),
- \( n \) is the number of moles,
- \( R \) is the universal gas constant in suitable units (we can use \( 0.0821 \, \text{L·atm/(mol·K)} \) for this calculation).
Rearranging the ideal gas law to solve for \( n \):
\[
n = \frac{PV}{RT}
\]
Substituting in the values:
\[
n = \frac{(3 \, \text{atm})(0.5 \, \text{L})}{(0.0821 \, \text{L·atm/(mol·K)})(270.15 \, \text{K})}
\]
Calculating the denominator:
\[
\text{denominator} = 0.0821 \times 270.15 \approx 22.196 \, \text{L·atm/(mol·K)}
\]
Now substituting back into the equation for \( n \):
\[
n = \frac{1.5 \, \text{atm·L}}{22.196 \, \text{L·atm/(mol·K)}} \approx 0.0674 \, \text{mol}
\]
Now, we can calculate the total translational kinetic energy of the gas using:
\[
KE_{\text{total}} = n \cdot KE_{\text{avg}} = n \cdot \frac{3}{2} RT
\]
Substituting the values into the kinetic energy equation:
\[
KE_{\text{total}} = 0.0674 \, \text{mol} \cdot \frac{3}{2} (8.314 \, \text{J/(mol·K)}) (270.15 \, \text{K})
\]
Calculating the average kinetic energy:
\[
KE_{\text{avg}} = \frac{3}{2} \cdot 8.314 \cdot 270.15 \approx \frac{3}{2} \cdot 2242.946 \approx 3364.42 \, \text{J/mol}
\]
Now substituting back to find the total kinetic energy:
\[
KE_{\text{total}} = 0.0674 \cdot 3364.42 \approx 226.43 \, \text{J}
\]
Thus, the total translational kinetic energy of 0.5 L of oxygen gas at -3 °C and 3 atm is approximately:
\[
\boxed{226.43 \, \text{J}}
\]