Question

At the surface of Venus the average temperature is a balmy 460∘C due to the greenhouse effect (global warming!), the pressure is 92 Earth-atmospheres, and the acceleration due to gravity is 0.894 gEarthgEarth. The atmosphere is nearly all CO2 (molar mass 44.0g/m) and the temperature remains remarkably constant. We shall assume that the temperature does not change at all with altitude.
a. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Earth-atmospheres.
b. What is the atmospheric pressure of 2.00 km above the surface of Venus? Express your answer in Venus-atmospheres.
c. What is the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km?

Answer 1

Answer:

a) 86 atm

b) 86 atm

c) 645 m/s

Explanation:

See attachment for calculations on how i arrived at the answer

Answer 2

The atmospheric pressure of 2.00 km above the surface of Venus is 86 atm and the root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km is 645 m/sec.

Given :

• At the surface of Venus, the average temperature is a balmy 460∘C due to the greenhouse effect (global warming!), the pressure is 92 Earth-atmospheres, and the acceleration due to gravity is 0.894 g.
• The atmosphere is nearly all CO2 (molar mass 44.0g/m) and the temperature remains remarkably constant.

a) and b) In order to determine the atmospheric pressure of 2.00 km above the surface of Venus using the formula given below:

$\rm P = P_0\times L^{\frac{Mg}{\rho T}}$   --- (1)

The value of the expression $\rm Mg/\rho T$ is given below:

$\rm \dfrac{Mg}{\rho T} = \dfrac{44\times 10^{-3}\times 9.8\times 10^3}{8.314\times 733}$

$\rm \dfrac{Mg}{\rho T}=0.02076$

Now, substitute the values of the known terms in the expression (1).

$\rm P=92\times L^{0.02076}$

P = 86 atm

c) The root-mean-square speed of the CO2 molecules at the surface of Venus and at an altitude of 2.00 km can be calculated as given below:

$\rm V_{rms}=\sqrt{\dfrac{3RT}{M}}$

Now, substitute the values of the known terms in the above formula.

$\rm V_{rms}=\sqrt{\dfrac{3\times 8.314\times 733}{44\times 10^{-3}}}$

Simplify the above expression.

$\rm V_{rms} = 645\;m/sec$

For more information, refer to the link given below:

https://brainly.com/question/7213287