Use R = 8.2 *10^-⁵ m³ . atm/mol . K and N_A = 6.02 * 10²³ mol^-¹. The approximate number of air molecules in a 1 m³ volume at room temperature (300 K) and atomospheric pressure is:

Question

Quizplus · Accepted Answer

We can use the ideal gas law to solve the problem: PV = nRT where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. We can rearrange this equation to solve for n: n = PV/RT Substituting in the given values: - P = 1 atm = 1.01325 × 10^5 Pa - V = 1 m^3 - R = 8.2 × 10^-2 atm/mol K - T = 300 K We can also use the value given for N_A to convert from moles to number of molecules: N_A = 6.02 × 10^23 molecules/mol Putting it all together: n = (1.01325 × 10^5 Pa)(1 m^3)/(8.2 × 10^-2 atm/mol K)(300 K) = 4.978 × 10^25 molecules Number of molecules = n × N_A = (4.978 × 10^25 molecules)(6.02 × 10^23 molecules/mol) ≈ 2.999 × 10^49 Rounding to the nearest power of 10, we get approximately 2.5 × 10^25, which is answer choice C.

Use R = 82 *10-5 M3

Use R = 82 *10^-⁵ M³