A 3.74-L cylinder contains 4.17 g of methane,CH₄,at a pressure of 3020 mmHg.What is the temperature of the gas? A)696°C B)423°C C)969°C D)2900°C E)129°C

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The ideal gas law, PV = nRT, can be used to find the temperature. First, convert the given values to appropriate units: pressure (P) = 3020 mmHg = 3020 mmHg * (1 atm / 760 mmHg) = 3.97 atm, mass of CH4 = 4.17 g, molar mass of CH4 = 16.04 g/mol (C = 12.01, H = 4 * 1.01), so moles (n) = 4.17 g / 16.04 g/mol = 0.26 mol. Volume (V) = 3.74 L, R = 0.0821 L·atm/(mol·K). Solve for T: T = PV / (nR) = (3.97 atm * 3.74 L) / (0.26 mol * 0.0821 L·atm/(mol·K)) = 696 K. Convert to °C: 696 K - 273 = 423°C.

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