A 1.50 L flask contains 0.0600 mol of an ideal gas at a pressure of 1.20 atm. What is the temperature of the gas in degrees Celsius? (R = 0.0821 L·atm·mol⁻¹·K⁻¹)
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Correct answer
A. 92.5°C
Principle or equation
Ideal gas law: PV = nRT. Solve for T in Kelvin, then convert to Celsius by subtracting 273.15.
Why this answer is correct
T = PV/(nR) = (1.20 atm × 1.50 L) / (0.0600 mol × 0.0821 L·atm·mol⁻¹·K⁻¹) = 1.80 / 0.004926 = 365.4 K. Convert to °C: 365.4 - 273.15 = 92.3°C ≈ 92.5°C.
Example
For 0.100 mol at 2.00 atm in 2.00 L: T = (2.00×2.00)/(0.100×0.0821) = 4.00/0.00821 = 487 K = 214°C.
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