A 2.00 L flask contains 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. -70.0
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) = (1.25 atm × 2.00 L) / (0.150 mol × 0.0821 L·atm·mol⁻¹·K⁻¹) = 2.50 / 0.012315 = 203.0 K. In Celsius: 203.0 - 273.15 = -70.15 °C, approximately -70.0 °C.
Example
For 0.100 mol at 1.00 atm in 2.00 L, T = (1.00×2.00)/(0.100×0.0821) = 243.6 K = -29.6 °C.
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