A 2.00 L flask contains 0.150 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
B. -23.0 °C
Principle or equation
Ideal gas law: PV = nRT. Solve for T in Kelvin, then convert to Celsius.
Why this answer is correct
T = PV/(nR) = (1.20 atm × 2.00 L) / (0.150 mol × 0.0821 L·atm·mol⁻¹·K⁻¹) = 2.40 / 0.012315 = 194.9 K. T(°C) = 194.9 - 273.15 = -78.3 °C ≈ -78.0 °C.
Example
If P=1.00 atm, V=1.00 L, n=0.0400 mol, T = 1.00/(0.0400×0.0821) = 304.5 K = 31.3 °C.
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