Answer:
Total pressure of the flask is 2.8999 atm.
Explanation:
Given data:
Volume of oxygen (O2) gas= 495 cm3
= 0.495 L (1 cm³ = 1 mL = 0.001 L)
Volume of nitrogen (N2) gas = 877 cm3
= 0.877 L (1 cm³ = 1 mL = 0.001 L)
volume of falsk = 536 cm3
= 0.536 L (1 cm³ = 1 mL = 0.001 L)
Temperature = 25 °C
T = (25°C + 273.15) K
= 298.15 K
Pressure = 114.7 kPa
= 114.700 Pa
Pressure (torr) = 114,700 / 101325
= 1.132 atm
Formula:
PV=nRT (ideal gas equation)
P = pressure
V = volume
R (gas constnt)= 0.0821 L.atm/K.mol
T = temperature
n = number of moles for both gases
Solution:
Firstly we will find the number of moles for oxygen and nitrogen gas.
For Oxygen:
n = PV / RT
n = 1.132 atm × 0.495 L / 0.0821 L.atm/K.mol × 298.15 K
= 0.560 / 24.47
= 0.0229 moles
For Nitrogen:
n = PV / RT
n = 1.132 atm × 0.877 / 0.0821 L.atm/K.mol × 298.15 K
n = 0.992 / 24.47
= 0.0406
Total moles = moles for oxygen gas + moles for nitrogen gas
= 0.0229 moles + 0.0406 moles
n = 0.0635 moles
Now put the values in formula
PV=nRT
P = nRT / V
P = 0.0635 × 0.0821 L.atm/K.mol × 298.15 K / 0.536 L
P = 1.554 / 0.536
P = 2.8999 atm
Total pressure in the flask is 2.8999 atm, while assuming the temperature constant.
The total pressure of gas in the flask at the given temperature is 3.2 atm.
The given parameters;
The number of moles of the nitrogen is calculated as follows;
[tex]PV = nRT\\\\n = \frac{PV}{RT} \\\\n = \frac{1.132 \times 0.877}{0.0821 \times (25 + 273)} \\\\n = 0.0406 \ mole[/tex]
The number of moles of oxygen is calculated as follows;
[tex]PV = nRT\\\\n = \frac{PV}{RT} \\\\n = \frac{1.132 \times 0.495}{0.0821 \times (25 + 273)} \\\\n = 0.0229 \ mole[/tex]
The total number of moles = 0.0406 mole + 0.0299 mole = 0.0705 mole.
The total pressure of gas in the flask is calculated as follows;
[tex]PV = nRT\\\\P = \frac{nRT}{V} \\\\P = \frac{0.0705 \times 0.0821 \times 298}{0.536} \\\\P = 3.2 \ atm[/tex]
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