Asked by Tesla
A sample of gas contains 6.25 × 10-3 mol in a 500.0 mL flask at 265°C. What is the pressure of the gas in kilopascals?
Formula: PV = nRT(R = 8.31 Lmr002-1.jpgkPa/molmr002-2.jpgK)
Which variables are given?
pressure
volume
temperature
moles
All Answers 1
Answered by
GPT-5 mini
AI
Which variables are given?
- Volume = 500.0 mL = 0.5000 L
- Temperature = 265°C = 265 + 273.15 = 538.15 K
- Moles = 6.25 × 10^-3 mol
- Pressure = unknown
Use PV = nRT → P = nRT/V
P = (6.25×10^-3 mol)(8.31 L·kPa·mol^-1·K^-1)(538.15 K) / (0.5000 L)
nRT ≈ 27.95 kPa·L → P ≈ 27.95 / 0.5000 = 55.9 kPa
Answer: 55.9 kPa (≈56.0 kPa with rounding).
- Volume = 500.0 mL = 0.5000 L
- Temperature = 265°C = 265 + 273.15 = 538.15 K
- Moles = 6.25 × 10^-3 mol
- Pressure = unknown
Use PV = nRT → P = nRT/V
P = (6.25×10^-3 mol)(8.31 L·kPa·mol^-1·K^-1)(538.15 K) / (0.5000 L)
nRT ≈ 27.95 kPa·L → P ≈ 27.95 / 0.5000 = 55.9 kPa
Answer: 55.9 kPa (≈56.0 kPa with rounding).
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