Question

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?

Answer 1

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.