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Answer :
Sure! Let's solve this question step-by-step.
---
Part a: Determining the Pressure of "Dry" Oxygen Gas
1. Identify the given values:
- Total atmospheric pressure, [tex]\( P_T = 100.2 \, \text{kPa} \)[/tex]
- Vapor pressure of water at [tex]\( 25^\circ \text{C} \)[/tex], [tex]\( P_{\text{H}_2\text{O}} = 3.2 \, \text{kPa} \)[/tex]
2. Calculate the pressure of the dry oxygen gas:
[tex]\[
P_{\text{gas}} = P_T - P_{\text{H}_2\text{O}} = 100.2 \, \text{kPa} - 3.2 \, \text{kPa} = 97.0 \, \text{kPa}
\][/tex]
---
Part b: Calculating the Moles of Oxygen Gas Produced
1. Identify additional given values:
- Volume of oxygen gas, [tex]\( V = 17.5 \, \text{mL} = 0.0175 \, \text{L} \)[/tex] (converted from mL to L)
- Universal gas constant, [tex]\( R = 0.08206 \, \frac{\text{L atm}}{\text{mol K}} \)[/tex]
- Temperature in Celsius, [tex]\( 25^\circ \text{C} \)[/tex]
2. Convert temperature to Kelvin:
[tex]\[
T = 25 + 273.15 = 298.15 \, \text{K}
\][/tex]
3. Convert pressure from kPa to atm:
[tex]\[
P_{\text{gas\_atm}} = \frac{97.0 \,\text{kPa}}{101.325 \,\text{kPa/atm}} \approx 0.957 \, \text{atm}
\][/tex]
4. Use the ideal gas law to calculate the number of moles of oxygen gas:
[tex]\[
PV = nRT \implies n = \frac{PV}{RT}
\][/tex]
[tex]\[
n = \frac{(0.957 \, \text{atm})(0.0175 \, \text{L})}{(0.08206 \, \frac{\text{L atm}}{\text{mol K}})(298.15 \, \text{K})}
\][/tex]
[tex]\[
n \approx 0.000685 \, \text{moles}
\][/tex]
So, the final answers are:
- The pressure of the dry oxygen gas is [tex]\( 97.0 \, \text{kPa} \)[/tex].
- The amount of moles of oxygen gas produced is approximately [tex]\( 0.000685 \, \text{moles} \)[/tex].
---
Part a: Determining the Pressure of "Dry" Oxygen Gas
1. Identify the given values:
- Total atmospheric pressure, [tex]\( P_T = 100.2 \, \text{kPa} \)[/tex]
- Vapor pressure of water at [tex]\( 25^\circ \text{C} \)[/tex], [tex]\( P_{\text{H}_2\text{O}} = 3.2 \, \text{kPa} \)[/tex]
2. Calculate the pressure of the dry oxygen gas:
[tex]\[
P_{\text{gas}} = P_T - P_{\text{H}_2\text{O}} = 100.2 \, \text{kPa} - 3.2 \, \text{kPa} = 97.0 \, \text{kPa}
\][/tex]
---
Part b: Calculating the Moles of Oxygen Gas Produced
1. Identify additional given values:
- Volume of oxygen gas, [tex]\( V = 17.5 \, \text{mL} = 0.0175 \, \text{L} \)[/tex] (converted from mL to L)
- Universal gas constant, [tex]\( R = 0.08206 \, \frac{\text{L atm}}{\text{mol K}} \)[/tex]
- Temperature in Celsius, [tex]\( 25^\circ \text{C} \)[/tex]
2. Convert temperature to Kelvin:
[tex]\[
T = 25 + 273.15 = 298.15 \, \text{K}
\][/tex]
3. Convert pressure from kPa to atm:
[tex]\[
P_{\text{gas\_atm}} = \frac{97.0 \,\text{kPa}}{101.325 \,\text{kPa/atm}} \approx 0.957 \, \text{atm}
\][/tex]
4. Use the ideal gas law to calculate the number of moles of oxygen gas:
[tex]\[
PV = nRT \implies n = \frac{PV}{RT}
\][/tex]
[tex]\[
n = \frac{(0.957 \, \text{atm})(0.0175 \, \text{L})}{(0.08206 \, \frac{\text{L atm}}{\text{mol K}})(298.15 \, \text{K})}
\][/tex]
[tex]\[
n \approx 0.000685 \, \text{moles}
\][/tex]
So, the final answers are:
- The pressure of the dry oxygen gas is [tex]\( 97.0 \, \text{kPa} \)[/tex].
- The amount of moles of oxygen gas produced is approximately [tex]\( 0.000685 \, \text{moles} \)[/tex].
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