Photo AI

130 cm³ of oxygen and 40 cm³ of nitrogen, each at 298 K and 100 kPa, were placed into an evacuated flask of volume 0.50 dm³ - AQA - A-Level Chemistry - Question 34 - 2017 - Paper 3

Question icon

Question 34

130-cm³-of-oxygen-and-40-cm³-of-nitrogen,-each-at-298-K-and-100-kPa,-were-placed-into-an-evacuated-flask-of-volume-0.50-dm³-AQA-A-Level Chemistry-Question 34-2017-Paper 3.png

130 cm³ of oxygen and 40 cm³ of nitrogen, each at 298 K and 100 kPa, were placed into an evacuated flask of volume 0.50 dm³. What is the pressure of the gas mixture... show full transcript

Worked Solution & Example Answer:130 cm³ of oxygen and 40 cm³ of nitrogen, each at 298 K and 100 kPa, were placed into an evacuated flask of volume 0.50 dm³ - AQA - A-Level Chemistry - Question 34 - 2017 - Paper 3

Step 1

Calculate Total Moles of Gases

96%

114 rated

Answer

To find the total pressure of the gas mixture, we first need to calculate the number of moles of each gas. Using the ideal gas law, we have:

For oxygen (O₂): nO2=PVRTn_{O_2} = \frac{PV}{RT} Where:

  • P = 100 kPa = 100,000 Pa
  • V = 130 cm³ = 0.000130 m³
  • R = 8.314 J/(mol·K)
  • T = 298 K

Substituting the values gives: nO2=100000×0.0001308.314×2980.0050 molesn_{O_2} = \frac{100000 \times 0.000130}{8.314 \times 298} \approx 0.0050 \text{ moles}

For nitrogen (N₂): nN2=PVRTn_{N_2} = \frac{PV}{RT} Where:

  • P = 100 kPa = 100,000 Pa
  • V = 40 cm³ = 0.000040 m³

Thus, substituting: nN2=100000×0.0000408.314×2980.0016 molesn_{N_2} = \frac{100000 \times 0.000040}{8.314 \times 298} \approx 0.0016 \text{ moles}

Step 2

Calculate Total Moles

99%

104 rated

Answer

Total moles of the gas mixture: ntotal=nO2+nN2=0.0050+0.0016=0.0066 molesn_{total} = n_{O_2} + n_{N_2} = 0.0050 + 0.0016 = 0.0066 \text{ moles}

Step 3

Calculate Pressure in Flask

96%

101 rated

Answer

Now, we want to find pressure in the flask at 298 K using volume of the flask (0.50 dm³ = 0.0005 m³):

Using the ideal gas law again: P=nRTVP = \frac{nRT}{V} Where:

  • n = 0.0066 moles
  • R = 8.314 J/(mol·K)
  • T = 298 K
  • V = 0.0005 m³

Substituting the values: P=0.0066×8.314×2980.000534.0 kPaP = \frac{0.0066 \times 8.314 \times 298}{0.0005} \approx 34.0 \text{ kPa}

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;