Two copper cylinders, immersed in a water tank at 46.3°C contain helium and nitrogen, respectively. The helium-filled cylinder has a volume twice as large as the nitrogen-filled cylinder.
a) Calculate the average kinetic energy of a helium molecule at 46.3°C.
b) Calculate the average kinetic energy of a nitrogen molecule at 46.3°C.
c) Determine the molar specific heat at constant volume (CV) and at constant pressure (Cp) for helium. Enter CV first. Give answer in J/mol·K.
d) Determine the molar specific heat at constant volume (CV) and at constant pressure (Cp) for nitrogen. Enter CV first. Give answer in J/mol·K.
e) Find γ for helium.
f) Find γ for nitrogen.
c) Cv=(3*R)/2=12.465 J/mol*K
Cp=((3+2)*R)/2=20.775 J/mol*K
d) Cv=(5*R)/2=20.775 J/mol*K
Cp=((5+2)*R)/2=29.085 J/mol*K
e) γHe=Cp/Cv=1.66
f) γN=Cp/Cv=1.4
if a mouse has a mass of 62g, what is its weight?
To calculate the average kinetic energy of molecules at a given temperature, we will use the formula:
KE = (3/2) * k * T
Where:
KE is the average kinetic energy
k is the Boltzmann constant (1.38 x 10^-23 J/K)
T is the temperature in Kelvin (K)
For part (a), we need to calculate the average kinetic energy of a helium molecule at 46.3°C.
Step 1: Convert temperature to Kelvin.
T = 46.3 + 273.15 = 319.45 K
Step 2: Plug the values into the formula.
KE = (3/2) * (1.38 x 10^-23 J/K) * (319.45 K)
≈ 3.19 x 10^-21 J
Therefore, the average kinetic energy of a helium molecule at 46.3°C is approximately 3.19 x 10^-21 J.
For part (b), we will follow the same steps to calculate the average kinetic energy of a nitrogen molecule at 46.3°C.
Step 1: Convert temperature to Kelvin.
T = 46.3 + 273.15 = 319.45 K
Step 2: Plug the values into the formula.
KE = (3/2) * (1.38 x 10^-23 J/K) * (319.45 K)
≈ 3.19 x 10^-21 J
Therefore, the average kinetic energy of a nitrogen molecule at 46.3°C is approximately 3.19 x 10^-21 J.
Moving on to parts (c) and (d), we need to determine the molar specific heat at constant volume (CV) and at constant pressure (Cp) for helium and nitrogen.
The molar specific heat at constant volume (CV) is given by:
CV = (3/2) * R
Where:
R is the gas constant (8.314 J/mol·K)
For helium:
CV(He) = (3/2) * (8.314 J/mol·K)
= 12.47 J/mol·K
For nitrogen:
CV(N2) = (3/2) * (8.314 J/mol·K)
= 12.47 J/mol·K
Therefore, the molar specific heat at constant volume for helium and nitrogen is both 12.47 J/mol·K.
For parts (e) and (f), we need to find γ (gamma), which is the ratio of specific heat at constant pressure (Cp) to specific heat at constant volume (CV).
γ = Cp / CV
For helium:
γ(He) = Cp(He) / CV(He)
To find Cp(He), we use the relationship Cp(He) = CV(He) + R.
Therefore:
γ(He) = (CV(He) + R) / CV(He)
= (12.47 J/mol·K + 8.314 J/mol·K) / 12.47 J/mol·K
≈ 1.67
For nitrogen:
γ(N2) = Cp(N2) / CV(N2)
Again, using Cp(N2) = CV(N2) + R:
γ(N2) = (CV(N2) + R) / CV(N2)
= (12.47 J/mol·K + 8.314 J/mol·K) / 12.47 J/mol·K
≈ 1.67
Therefore, the value of γ for both helium and nitrogen is approximately 1.67.