Question
A 2.30 L container holds 24.00 grams of oxygen gas at 108.5 kPa at 22.00oC. What will be the final volume of the container if the pressure increases by 4.20 kPa, the amount increases to 1.85 moles, and the temperature increases to 52.00oC?
Solve using combin gas law
Solve using combin gas law
Answers
GPT-4o mini
To find the final volume of the container after the changes in pressure, amount of gas (in moles), and temperature, we can use the combined gas law, which is expressed as:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Where:
- \(P\) = pressure
- \(V\) = volume
- \(T\) = temperature (in Kelvin)
First, we need to convert all the values into the appropriate units:
### Initial Conditions
1. **Initial pressure (\(P_1\))**: \(108.5 \, \text{kPa}\)
2. **Initial volume (\(V_1\))**: \(2.30 \, \text{L}\)
3. **Initial temperature (\(T_1\))**: Convert \(22.00^\circ C\) to Kelvin:
\[
T_1 = 22.00 + 273.15 = 295.15 \, \text{K}
\]
### Final Conditions
1. **Final pressure (\(P_2\))**:
\[
P_2 = 108.5 \, \text{kPa} + 4.20 \, \text{kPa} = 112.7 \, \text{kPa}
\]
2. **Final volume (\(V_2\))**: (unknown, we need to solve for this)
3. **Final temperature (\(T_2\))**: Convert \(52.00^\circ C\) to Kelvin:
\[
T_2 = 52.00 + 273.15 = 325.15 \, \text{K}
\]
Now we can set up our equation using the combined gas law:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Substituting in our known values:
\[
\frac{(108.5 \, \text{kPa})(2.30 \, \text{L})}{295.15 \, \text{K}} = \frac{(112.7 \, \text{kPa})(V_2)}{325.15 \, \text{K}}
\]
Now, we can cross multiply to solve for \(V_2\):
\[
(108.5 \, \text{kPa})(2.30 \, \text{L})(325.15 \, \text{K}) = (112.7 \, \text{kPa})(V_2)(295.15 \, \text{K})
\]
Calculating the left side:
\[
(108.5)(2.30)(325.15) = 78912.3375 \, \text{kPa} \cdot \text{L} \cdot \text{K}
\]
Calculating the right side (keeping it in equation form for now):
\[
(112.7)(V_2)(295.15) = 33378.905V_2 \, \text{kPa} \cdot \text{L} \cdot \text{K}
\]
Now set the two sides equal:
\[
78912.3375 = 33378.905V_2
\]
Now solve for \(V_2\):
\[
V_2 = \frac{78912.3375}{33378.905} \approx 2.36 \, \text{L}
\]
Thus, the final volume of the container is approximately **2.36 liters**.
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Where:
- \(P\) = pressure
- \(V\) = volume
- \(T\) = temperature (in Kelvin)
First, we need to convert all the values into the appropriate units:
### Initial Conditions
1. **Initial pressure (\(P_1\))**: \(108.5 \, \text{kPa}\)
2. **Initial volume (\(V_1\))**: \(2.30 \, \text{L}\)
3. **Initial temperature (\(T_1\))**: Convert \(22.00^\circ C\) to Kelvin:
\[
T_1 = 22.00 + 273.15 = 295.15 \, \text{K}
\]
### Final Conditions
1. **Final pressure (\(P_2\))**:
\[
P_2 = 108.5 \, \text{kPa} + 4.20 \, \text{kPa} = 112.7 \, \text{kPa}
\]
2. **Final volume (\(V_2\))**: (unknown, we need to solve for this)
3. **Final temperature (\(T_2\))**: Convert \(52.00^\circ C\) to Kelvin:
\[
T_2 = 52.00 + 273.15 = 325.15 \, \text{K}
\]
Now we can set up our equation using the combined gas law:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Substituting in our known values:
\[
\frac{(108.5 \, \text{kPa})(2.30 \, \text{L})}{295.15 \, \text{K}} = \frac{(112.7 \, \text{kPa})(V_2)}{325.15 \, \text{K}}
\]
Now, we can cross multiply to solve for \(V_2\):
\[
(108.5 \, \text{kPa})(2.30 \, \text{L})(325.15 \, \text{K}) = (112.7 \, \text{kPa})(V_2)(295.15 \, \text{K})
\]
Calculating the left side:
\[
(108.5)(2.30)(325.15) = 78912.3375 \, \text{kPa} \cdot \text{L} \cdot \text{K}
\]
Calculating the right side (keeping it in equation form for now):
\[
(112.7)(V_2)(295.15) = 33378.905V_2 \, \text{kPa} \cdot \text{L} \cdot \text{K}
\]
Now set the two sides equal:
\[
78912.3375 = 33378.905V_2
\]
Now solve for \(V_2\):
\[
V_2 = \frac{78912.3375}{33378.905} \approx 2.36 \, \text{L}
\]
Thus, the final volume of the container is approximately **2.36 liters**.