Question 01:
To find the average velocity of nitrogen molecules, we can use the formula for average velocity which is given by:
v = sqrt(8kT / πm)
where v is the average velocity, k is the Boltzmann constant (1.38 x 10^-23 J/K), T is the temperature in Kelvin, and m is the molar mass of nitrogen gas (28 g/mol).
First, we need to convert the mass of nitrogen gas from grams to moles by dividing it by the molar mass:
Number of Moles = 2.0g / 28 g/mol = 0.0714 mol
Next, we need to convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = 270°C + 273.15 = 543.15 K
Now we can substitute these values into the formula to find the average velocity:
v = sqrt(8 * 1.38 x 10^-23 J/K * 543.15 K / π * 28 g/mol)
v ≈ 515.04 m/s
To find the temperature for 2.0g of Helium gas with an average velocity of 20 m/s, we can rearrange the formula to solve for temperature:
T = (v^2 * πm) / (8k)
Let's plug in the values:
T = (20 m/s)^2 * π * 4 g/mol / (8 * 1.38 x 10^-23 J/K)
T ≈ 230,053 K
As for plotting the Cv v/s T curve based on the Kinetic theory, I would need more specific information about the specific heat capacities (Cv) for nitrogen and helium gases over a range of temperatures to accurately create the schematic plot.
Question 02:
To find the total entropy change of the ice-lake system, we can use the formula for entropy change given by:
ΔS = Q / T
where ΔS is the entropy change, Q is the heat transferred, and T is the temperature in Kelvin.
First, we need to determine the heat transferred during the melting process. The heat transferred is given by:
Q = ml
where m is the mass of the ice and l is the latent heat of fusion of ice (also known as the heat of fusion). The latent heat of fusion for ice is approximately 334 J/g.
For the first scenario where the ice is introduced at 27°C:
Q = 200g * 334 J/g = 66,800 J
Next, we need to convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = 27°C + 273.15 = 300.15 K
Now we can substitute these values into the entropy change formula:
ΔS = 66,800 J / 300.15 K
ΔS ≈ 222.57 J/K
For the second scenario where the ice is introduced at 0.20°C:
Q = 200g * 334 J/g = 66,800 J
Temperature in Kelvin = 0.20°C + 273.15 = 273.35 K
ΔS = 66,800 J / 273.35 K
ΔS ≈ 244.57 J/K
Based on the results, we can conclude that the entropy change tends to be irreversible, meaning that there is an increase in the total entropy of the system as the ice is introduced to the lake. This is because the process involves heat transfer from a higher temperature (the lake) to a lower temperature (the ice), which is a spontaneous process that tends to increase the entropy.