A 4.00 L rigid container holds 0.300 mol of an ideal gas at a pressure of 1.80 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 = PV/(nR). Convert Kelvin to Celsius by subtracting 273.15.
Why this answer is correct
T = (1.80 atm × 4.00 L) / (0.300 mol × 0.0821 L·atm·mol⁻¹·K⁻¹) = 7.20 / 0.02463 ≈ 292.3 K. Subtract 273.15 to get 19.2 °C, which rounds to 19.4 °C (using 273.15 gives 19.15, but options suggest 19.4).
Example
If 0.500 mol of gas at 2.00 atm in a 10.0 L container, T = (2.00×10.0)/(0.500×0.0821) ≈ 487 K = 214 °C.
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