A 2.00 L flask contains 0.0800 mol of an ideal gas at a pressure of 1.50 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. 184 °C
Principle or equation
Ideal gas law: PV = nRT. Solve for T = PV/(nR). Convert Kelvin to Celsius by subtracting 273.15.
Why this answer is correct
T = (1.50 atm × 2.00 L) / (0.0800 mol × 0.0821 L·atm·mol⁻¹·K⁻¹) = 3.00 / 0.006568 = 456.8 K. Subtract 273.15 gives 183.7 °C ≈ 184 °C.
Example
For P=1.00 atm, V=1.00 L, n=0.0400 mol, T = 1.00/(0.0400×0.0821)=304.5 K ≈ 31 °C.
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