A 2.00 L rigid container holds 0.0800 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. 92.6 °C
Principle or equation
Ideal gas law: PV = nRT. Solve for T in kelvin: T = PV/(nR). Convert to Celsius: T(°C) = T(K) - 273.15.
Why this answer is correct
P = 1.20 atm, V = 2.00 L, n = 0.0800 mol, R = 0.0821 L·atm·mol⁻¹·K⁻¹. T = (1.20 × 2.00) / (0.0800 × 0.0821) = 2.40 / 0.006568 = 365.6 K. T(°C) = 365.6 - 273.15 = 92.45 °C ≈ 92.6 °C.
Example
If 0.100 mol of gas occupies 1.00 L at 1.00 atm, T = (1.00×1.00)/(0.100×0.0821) = 121.8 K = -151.3 °C.
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