Question 13 to 14.
13.How many molecules are there 70cm^{3} of benzene? For benzene, \rho=0.88g/cm^{3} and M = 78kg/kmol.
4.7\times10^{-26} . 3.0\times10^{23} c. 4.8\times10^{23}d.3.2\times10^{23}
14. At what pressure will the mean-free-path be 50\times10^{-2}m for spherical molecules of radius 3.0\times10^{-10}m?
Assume an ideal gas at 20^{\circ}C.
a. 5.1\times10^{-3}Pa b. 6.2\times10^{-3}Pac. 4.8\times10^{-3}Pa=2.6\times10^{-3}Pa
15. A gas does 5J of work while expanding adiabatically. What is the change in internal energy? a. 6J b. -6J c. 5/ d. -5/
13. To find the number of molecules in 70cm^3 of benzene, we need to use the formula:
Number of molecules = (Volume in cm^3 / molar volume) * Avogadro's number
First, let's convert the given volume of 70cm^3 to liters:
70cm^3 = 70/1000 = 0.07 L
Next, we need to calculate the molar volume of benzene using the given density and molar mass:
Molar volume = (Molar mass / Density) * 1000
Molar mass of benzene = 78kg/kmol = 78,000g/kmol
Density of benzene = 0.88g/cm^3
Molar volume = (78,000 / 0.88) * 1000 = 88,636.36 cm^3/mol
Now, let's calculate the number of molecules:
Number of molecules = (0.07 / 88,636.36) * 6.022 × 10^23
Simplifying this equation, we get:
Number of molecules ≈ 4.7 × 10^23
Therefore, the answer is option a. 4.7 × 10^23.
14. The mean-free-path (λ) can be determined using the kinetic theory of gases equation:
λ = (k * T) / (√2 * π * d^2 * P)
Where:
k = Boltzmann constant = 1.38 × 10^-23J/K
T = temperature in Kelvin (20°C = 293K)
d = diameter of molecules = 2 * 3.0 × 10^-10m = 6.0 × 10^-10m
To find the pressure (P) at which the mean-free-path is 50 × 10^-2m, we can rearrange the equation:
P = (k * T) / (√2 * π * d^2 * λ)
Plugging in the given values, we get:
P = (1.38 × 10^-23 * 293) / (√2 * π * (6.0 × 10^-10)^2 * 50 × 10^-2)
Simplifying this equation, we get:
P ≈ 6.2 × 10^-3Pa
Therefore, the answer is option b. 6.2 × 10^-3Pa.
15. In an adiabatic process, there is no heat transfer, meaning that the change in internal energy (ΔU) is equal to the work done (W).
Given that the gas does 5J of work, we can conclude that the change in internal energy is also 5J.
Therefore, the answer is option c. 5J.