A 2.50 L rigid container holds 0.150 mol of an ideal gas at a pressure of 1.25 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. -19.4°C
Principle or equation
Ideal gas law: PV = nRT. Solve for T in kelvin, then convert to Celsius: T(°C) = T(K) - 273.15.
Why this answer is correct
P = 1.25 atm, V = 2.50 L, n = 0.150 mol. T = PV/(nR) = (1.25 * 2.50) / (0.150 * 0.0821) = 3.125 / 0.012315 = 253.8 K. Convert: 253.8 - 273.15 = -19.35°C ≈ -19.4°C.
Example
For 0.300 mol in 4.00 L at 1.80 atm, T = (1.80*4.00)/(0.300*0.0821) = 292.3 K = 19.1°C.
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