the particles in which sample of LiCl(s) have the same average kinetic energy as the particles in a 2.0-mole sample ph H2O(l) at 25 degrees?
To find the particles in a sample of LiCl(s) with the same average kinetic energy as a 2.0-mole sample of H2O(l) at 25 degrees Celsius, we need to use the equation for the average kinetic energy of an ideal gas:
Ek = (3/2) * k * T
Where:
Ek is the average kinetic energy of the particles
k is Boltzmann's constant (1.38 × 10^-23 J/K)
T is the temperature in Kelvin
First, we need to convert the temperature from Celsius to Kelvin:
T(K) = T(°C) + 273.15
T = 25°C + 273.15 = 298.15 K
Now, we can calculate the average kinetic energy for both substances.
For the 2.0-mole sample of H2O(l):
Ek(H2O) = (3/2) * k * T
Ek(H2O) = (3/2) * (1.38 × 10^-23 J/K) * (298.15 K)
For the sample of LiCl(s), we need to consider that this is a solid, not an ideal gas. Solid particles have different modes of kinetic energy, such as vibrational and rotational energies, in addition to translational kinetic energy. Therefore, we can't directly compare the average kinetic energy of LiCl(s) to that of H2O(l).
However, if we assume that we only consider the translational kinetic energy for both substances, we can use the formula above to estimate the average kinetic energy.
Since the molar mass of H2O is 18.015 g/mol and the molar mass of LiCl is 42.39 g/mol, we can calculate the ratio of the molar masses:
Ratio of molar masses = M(H2O) / M(LiCl)
= 18.015 g/mol / 42.39 g/mol
≈ 0.425
Now, we can calculate the average kinetic energy for LiCl(s) based on the calculated ratio:
Ek(LiCl) = Ek(H2O) * Ratio of molar masses
= (3/2) * (1.38 × 10^-23 J/K) * (298.15 K) * 0.425
By solving this equation, you can find the average kinetic energy of the particles in a sample of LiCl(s) with the same average kinetic energy as the particles in a 2.0-mole sample of H2O(l) at 25 degrees Celsius.