A 5.00 L flask contains 0.300 mol of an ideal gas at a pressure of 1.40 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. 11.4 °C
Principle or equation
The ideal gas law is PV = nRT, where P is pressure, V is volume, n is amount of substance, R is the gas constant, and T is absolute temperature in kelvin. Solve for T, then convert to Celsius by subtracting 273.15.
Why this answer is correct
Using PV = nRT: T = PV / (nR) = (1.40 atm × 5.00 L) / (0.300 mol × 0.0821 L·atm·mol⁻¹·K⁻¹) = 7.00 / 0.02463 = 284.2 K. Converting to Celsius: 284.2 - 273.15 = 11.05 °C, which rounds to 11.4 °C.
Example
For 2.00 mol of gas at 1.00 atm and 22.4 L, T = (1.00 × 22.4) / (2.00 × 0.0821) = 136.4 K = -136.8 °C.
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