The particles of which sample of LiCl(s) have the same average kinetic energy as the particles in a 2.0 mole sample of NH3(g) at 25 degrees celcius?


a) 1.0 mole @ 75 degrees celcius
b) 12.0 mole @ 50 degrees celcius
c) 3.0 mole @ 25 degrees celcius
d) 4.0 mole @ 0 degrees celcius

What is the avg KE of LiCl?

What is the avg KE of a, b, c, d?

To compare the average kinetic energy between samples of different substances, we can consider the relationship between temperature and kinetic energy. The average kinetic energy of particles in a sample is directly proportional to the temperature.

The formula to calculate the average kinetic energy (E) is:

E = (3/2) * k * T

Where:
- E is the average kinetic energy
- k is Boltzmann's constant (1.38 * 10^-23 J/K)
- T is the temperature in Kelvin

To determine which sample of LiCl(s) has the same average kinetic energy as the particles in a 2.0 mole sample of NH3(g), we need to calculate the average kinetic energy for both substances.

First, let's calculate the average kinetic energy for the 2.0 mole sample of NH3(g) at 25 degrees Celsius (298 Kelvin):

E_NH3 = (3/2) * k * T
= (3/2) * (1.38 * 10^-23 J/K) * 298 K
= 6.18 * 10^-21 J

Now, let's calculate the average kinetic energy for each sample of LiCl(s) given in the options:

a) 1.0 mole @ 75 degrees Celsius (348 Kelvin):
E_LiCl_a = (3/2) * k * T
= (3/2) * (1.38 * 10^-23 J/K) * 348 K
= 9.06 * 10^-21 J

b) 12.0 mole @ 50 degrees Celsius (323 Kelvin):
E_LiCl_b = (3/2) * k * T
= (3/2) * (1.38 * 10^-23 J/K) * 323 K
= 8.42 * 10^-21 J

c) 3.0 mole @ 25 degrees Celsius (298 Kelvin):
E_LiCl_c = (3/2) * k * T
= (3/2) * (1.38 * 10^-23 J/K) * 298 K
= 6.18 * 10^-21 J

d) 4.0 mole @ 0 degrees Celsius (273 Kelvin):
E_LiCl_d = (3/2) * k * T
= (3/2) * (1.38 * 10^-23 J/K) * 273 K
= 5.33 * 10^-21 J

Comparing the calculated values, we can see that the sample with the same average kinetic energy as the 2.0 mole sample of NH3(g) at 25 degrees Celsius is option c) 3.0 mole @ 25 degrees Celsius.