To solve the problem of comparing the volumes of nitrogen gas and oxygen gas at standard temperature and pressure (STP), we will follow these steps:
### Step 1: Understand the Conditions
We are given:
1. **Nitrogen gas (N₂)**: Volume = 1.2 L, Temperature = 25°C, Pressure = 748 mm Hg
2. **Oxygen gas (O₂)**: Volume = 1.25 L at STP
STP conditions are defined as:
- Temperature = 0°C (273 K)
- Pressure = 760 mm Hg
### Step 2: Convert Temperature to Kelvin
For nitrogen gas:
- Temperature in Kelvin = 25°C + 273 = 298 K
### Step 3: Use the Ideal Gas Law
We will use the formula derived from the ideal gas law, which relates pressure, volume, and temperature:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Where:
- \(P_1\), \(V_1\), \(T_1\) are the initial pressure, volume, and temperature of nitrogen.
- \(P_2\), \(V_2\), \(T_2\) are the standard pressure, volume, and temperature.
### Step 4: Assign Known Values
For nitrogen gas:
- \(P_1 = 748 \, \text{mm Hg}\)
- \(V_1 = 1.2 \, \text{L}\)
- \(T_1 = 298 \, \text{K}\)
For STP:
- \(P_2 = 760 \, \text{mm Hg}\)
- \(T_2 = 273 \, \text{K}\)
- \(V_2\) is what we want to find.
### Step 5: Rearrange the Ideal Gas Law
Rearranging the equation to solve for \(V_2\):
\[
V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}
\]
### Step 6: Substitute the Values
Now substituting the known values into the equation:
\[
V_2 = 1.2 \, \text{L} \times \frac{748 \, \text{mm Hg}}{760 \, \text{mm Hg}} \times \frac{273 \, \text{K}}{298 \, \text{K}}
\]
### Step 7: Calculate \(V_2\)
Calculating each part:
1. Calculate the pressure ratio:
\[
\frac{748}{760} \approx 0.9842
\]
2. Calculate the temperature ratio:
\[
\frac{273}{298} \approx 0.9164
\]
3. Now, multiply these ratios with the volume:
\[
V_2 = 1.2 \times 0.9842 \times 0.9164 \approx 1.081 \, \text{L}
\]
### Step 8: Compare the Volumes
Now we have:
- Volume of nitrogen gas at STP = 1.081 L
- Volume of oxygen gas at STP = 1.25 L
### Conclusion
Since 1.25 L (oxygen) > 1.081 L (nitrogen), the oxygen gas has a greater volume at STP.
