To solve the problem of calculating the volume of moist nitrogen gas at Normal Temperature and Pressure (NTP), we will follow these steps:
### Step 1: Identify Given Values
- Volume of moist nitrogen, \( V_1 = 38.0 \, \text{mL} \)
- Temperature, \( T_1 = 27^\circ C \)
- Pressure, \( P_1 = 0.98 \, \text{bar} \)
- Aqueous tension at \( 27^\circ C = 0.035 \, \text{bar} \)
### Step 2: Calculate the Pressure of the Dry Gas
To find the pressure of the dry nitrogen gas, we need to subtract the aqueous tension from the total pressure:
\[
P_{\text{dry}} = P_1 - \text{Aqueous Tension} = 0.98 \, \text{bar} - 0.035 \, \text{bar} = 0.945 \, \text{bar}
\]
### Step 3: Convert Temperature to Kelvin
Convert the temperature from Celsius to Kelvin:
\[
T_1 = 27^\circ C + 273 = 300 \, \text{K}
\]
The normal temperature \( T_2 \) at NTP is:
\[
T_2 = 0^\circ C + 273 = 273 \, \text{K}
\]
### Step 4: Identify Normal Pressure
The normal pressure \( P_2 \) at NTP is:
\[
P_2 = 1.013 \, \text{bar}
\]
### Step 5: Use the Ideal Gas Law
We will use the ideal gas law in the form of:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Rearranging for \( V_2 \):
\[
V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}
\]
### Step 6: Substitute the Values
Substituting the known values into the equation:
\[
V_2 = \frac{(0.945 \, \text{bar}) \times (38.0 \, \text{mL}) \times (273 \, \text{K})}{(1.013 \, \text{bar}) \times (300 \, \text{K})}
\]
### Step 7: Calculate \( V_2 \)
Calculating \( V_2 \):
\[
V_2 = \frac{(0.945) \times (38.0) \times (273)}{(1.013) \times (300)}
\]
\[
V_2 = \frac{(945 \times 38 \times 273)}{(1.013 \times 300)}
\]
\[
V_2 = \frac{(945 \times 38 \times 273)}{303.9}
\]
Calculating the numerator:
\[
945 \times 38 \times 273 \approx 944,205
\]
Calculating the denominator:
\[
1.013 \times 300 \approx 303.9
\]
Now, dividing:
\[
V_2 \approx \frac{944,205}{303.9} \approx 31.06 \, \text{mL}
\]
### Final Result
The volume of nitrogen gas at NTP is approximately:
\[
V_2 \approx 32.26 \, \text{mL}
\]
