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 gEarth. The atmosphere is nearly all CO2 (molar mass 44.0g/mol) and the temperature remains remarkably constant. We shall assume that the temperature does not change at all with altitude.

Part A

What is the atmospheric pressure 2.00 km above the surface of Venus? Express your answer in Earth-atmospheres.

Express your answer using two significant figures.

P = Earth atm

Part B

What is the atmospheric pressure 2.00 km above the surface of Venus? Express your answer in Venus-atmospheres.

Express your answer using two significant figures.

P =   Venus atm

Part 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 ?

vrms =   m/s

Answer 1

Part A\&B:

For constant temperature,the variation of pressure with altitudeis \(\rho=\rho_{0} e^{\frac{-M g y}{R T}}\)

\(v_{r m}=\sqrt{\frac{3 R T}{M}}\)

\(g=9.8 \mathrm{~m} / \mathrm{s}^{2}, T=460^{\circ} \mathrm{C}=733 \mathrm{~K}, M=44.0 \mathrm{~g} / \mathrm{mol}=44 \times 10^{-3} \mathrm{~kg} / \mathrm{mol}\)

\(\frac{M g y}{R T}=\frac{\left(44 \times 10^{-3}\right)(0.894)(9.80)\left(2 \times 10^{3}\right)}{(8.314)(733)}=0.1265\)

\(\rho=\rho_{0} e^{\frac{-M g v}{R T}}=(92) e^{-0.1265}=81.068 \mathrm{~atm}\)

The pressure is \(81.068\) earth-atmospheres \(\frac{81.068}{92}=0.8811\) Venus atmospheres.

Part C:

\(v_{r \pi}=\sqrt{\frac{3 R T}{M}}=\sqrt{\frac{3 \times 8.314 \times 733}{44 \times 10^{-3}}}=644.6 \mathrm{~m} / \mathrm{s} \approx 645 \mathrm{~m} / \mathrm{s}\)

\(v_{r w c}\) has this value both at the surface and at an altitude of \(2.0 \mathrm{~km}\)