a) Determine the mean molar mass of the atmosphere of Venus, which consists of 95% CO2 and 5% N2 by volume.
b) What is the corresponding gas constant?
c) The mean surface temperature T on Venus is a scorching 740K as compared to only 288K for Earth; the surface pressure is 90 times that on Earth. By what factor is the density of the near-surface Venusian atmosphere greater or less than that of Earth?
Solution.
a)
"M = \\frac{1}{(\\frac{w(CO2)}{M(CO2)}+\\frac{w(N2)}{M(N2)})}"
"M = \\frac{1}{\\frac{0.95}{44}+\\frac{0.05}{28}} = 42.78 \\ kg\/kmol"
b)
"R = \\frac{8314}{M} = 194.34 \\ \\frac{J}{kg \\times K}"
c)
"R(E) = \\frac{8314}{29} = 287"
"\\frac{\\rho(E)}{\\rho(V)} = \\frac{p(E)*M(E)*R(V)*T(V)}{R(E)*T(E)*p(V)*M(V)} = 0.013"
Answer:
a)
M = 42.78 kg/kmol
b)
"R = 194.34 \\ \\frac{J}{kg \\times K}"
c)
"\\frac{\\rho(E)}{\\rho(V)} = 0.013"
The density of the atmosphere of Venus is 76.92 times greater than that of the Earth's atmosphere.
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