Question

Find the total number of collisions between molecules in 1.00 s in 1.00 L of nitrogen gas at standard temperature and pressure (0°C, 1.00 atm). Use 1.88×10⁻10m as the effective radius of a nitrogen molecule. (The number of collisions per second is the reciprocal of the collision time.) Keep in mind that each collision involves two molecules, so if one molecule collides once in a certain period of time, the collision of the molecule it hit cannot be counted.

a) 4.27 × 10²5 collisions
b) 1.78 × 10²6 collisions
c) 3.21 × 10²6 collisions
d) 6.15 × 10²6 collisions

Answer 1

To find the total number of collisions between molecules in 1.00 s in 1.00 L of nitrogen gas at standard temperature and pressure,

calculate the number of molecules using the ideal gas law and the effective radius of a nitrogen molecule. Then, divide the reciprocal of the collision time by 2 to get the total number of collisions between molecules.

Step-by-step explanation:

To find the total number of collisions between molecules in 1.00 s in 1.00 L of nitrogen gas at standard temperature and pressure (0°C, 1.00 atm), we can use the formula:

N = (N/V) * V = (p/kB T) * V

where N is the total number of molecules, V is the volume of the gas, p is the pressure, kB is Boltzmann's constant, and T is the temperature. We know the effective radius of a nitrogen molecule is 1.88×10⁻¹⁰m. Using the ideal gas law, we can calculate the value of N/V and then multiply it by the volume of the gas to find N.

After calculating the value of N, we can find the number of collisions by taking the reciprocal of the collision time, which is given as 1.00 s.

Since each collision involves two molecules, we need to divide the total number of collisions by 2 to get the final result. Plugging in the values, the total number of collisions between molecules in 1.00 s in 1.00 L of nitrogen gas at standard temperature and pressure is approximately 1.78 × 10²⁶ collisions.