A 4.00 L flask contains 0.250 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. 19.5 °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) = (1.50 atm × 4.00 L) / (0.250 mol × 0.0821 L·atm·mol⁻¹·K⁻¹) = 6.00 / 0.020525 = 292.3 K. Convert to °C: 292.3 - 273.15 = 19.2 °C, approximately 19.5 °C.
Example
For 0.500 mol at 2.00 atm in 10.0 L, T = (2.00 × 10.0)/(0.500 × 0.0821) = 487 K = 214 °C.
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