A 2.50 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
A. -29°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) = (1.20 atm × 2.50 L) / (0.150 mol × 0.0821 L·atm·mol⁻¹·K⁻¹) = 3.00 / (0.012315) ≈ 243.6 K. Convert to Celsius: 243.6 - 273.15 = -29.55°C ≈ -29°C.
Example
For 0.200 mol in 4.00 L at 1.00 atm, T = (1.00 × 4.00)/(0.200 × 0.0821) = 243.6 K = -29°C.
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