A 1.80 L flask contains 0.120 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. 1.0 °C
Principle or equation
The ideal gas law is PV = nRT. Solve for T in kelvins, then convert to Celsius by subtracting 273.15.
Why this answer is correct
T = PV/(nR) = (1.50 atm × 1.80 L) / (0.120 mol × 0.0821 L·atm·mol⁻¹·K⁻¹) = 2.70 / 0.009852 = 274.1 K. In Celsius: 274.1 - 273.15 = 1.0 °C.
Example
For 0.250 mol in 3.50 L at 1.60 atm, T = (1.60 × 3.50)/(0.250 × 0.0821) = 273 K = 0 °C.
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