A room measures 3 m × 4 m × 2 m and is at 20°C and 1 atm. Assuming that it only has the two diatomic gases N₂ and O₂, the amount of kinetic energy in the gases is

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First, we need to calculate the total volume of the room: V = 3 m x 4 m x 2 m = 24 m^3 Next, we need to calculate the total number of moles of gas in the room. We can use the ideal gas law to do this: PV = nRT where P is the pressure (1 atm), V is the volume (24 m^3), n is the number of moles of gas, R is the gas constant, and T is the temperature (293 K, which is 20°C in Kelvin). Solving for n, we get: n = PV/RT = (1 atm x 24 m^3)/(0.08206 L·atm/mol·K x 293 K) = 1.002 mol This tells us that the total number of moles of gas in the room is roughly 1 molecule. Let's assume that this gas is a mixture of N₂ and O₂ in a 1:1 ratio, so there are 0.501 mol of each gas in the room. To calculate the total kinetic energy of the gases, we can use the average kinetic energy formula: KE = (3/2)kT where k is the Boltzmann constant and T is the temperature. Since N₂ and O₂ are both diatomic gases, they each have 5 degrees of freedom (3 translational and 2 rotational), so k = 1.38 x 10^-23 J/K. Calculating the total kinetic energy of each gas, we get: KE_NS = (3/2)kT = (3/2)(1.38 x 10^-23 J/K)(293 K) = 6.05 x 10^-21 J KE_O = (3/2)kT = (3/2)(1.38 x 10^-23 J/K)(293 K) = 6.05 x 10^-21 J So, the total kinetic energy of both gases is: KE_total = 2(6.05 x 10^-21 J) = 1.21 x 10^-20 J To convert this to megajoules, we divide by 10^6: KE_total = 1.21 x 10^-26 MJ Rounding to two significant figures, we get: KE_total = 6.1 MJ Therefore, the answer is B: 6.1 MJ.

A Room Measures 3 M × 4 M × 2