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Chapter 18: Problem 54

At the surface of Venus the average temperature is a balmy 460\(^\circ\)C due to the greenhouse effect (global warming!), the pressure is 92 earth-atmospheres, and the acceleration due to gravity is 0.894\(g{_e}{_a}{_r}{_t}{_h}\). The atmosphere is nearly all CO\(_2\) (molar mass 44.0 g/mol), and the temperature remains remarkably constant. Assume that the temperature does not change with altitude. (a) What is the atmospheric pressure 1.00 km above the surface of Venus? Express your answer in Venus-atmospheres and earth-atmospheres. (b) What is the root-mean-square speed of the CO\(_2\) molecules at the surface of Venus and at an altitude of 1.00 km?

Short Answer

Expert verified
(a) 86.6 Venus-atm; (b) 472.4 m/s at both altitudes.

Step by step solution

01

Understanding Pressure Change with Altitude

For Venus, we need to calculate the atmospheric pressure at 1.00 km altitude. We start with the barometric formula for a non-ideal gas where the temperature remains constant:\[P(h) = P_0 \, e^{-\frac{Mgh}{RT}}\]where:- \(P(h)\) is the pressure at height \(h\)- \(P_0 = 92\) atm (surface pressure of Venus)- \(M = 0.044\) kg/mol (molar mass of CO\(_2\))- \(g = 0.894 \, g_{earth} = 0.894 \times 9.81 \, m/s^2\)- \(h = 1000\) m- \(R = 8.314 \, J/(mol \cdot K)\) (universal gas constant)- \(T = 460 + 273.15 = 733.15 \, K\) (temperature in Kelvin)First, we find \(Mg/RT\).
02

Calculate Mg/RT

Let's calculate \(\frac{Mg}{RT}\):\[\frac{Mg}{RT} = \frac{0.044 \times (0.894 \times 9.81)}{8.314 \times 733.15}\]Evaluating this expression gives us:\[\frac{Mg}{RT} \approx 5.98 \times 10^{-5} \, m^{-1}\]
03

Evaluate the Exponential

Now evaluate the exponential term \(e^{-\frac{Mgh}{RT}}\) using \(h = 1000\) m:\[e^{-\frac{Mgh}{RT}} = e^{-(5.98 \times 10^{-5}) \times 1000}\]This results in:\[e^{-0.0598} \approx 0.942\]
04

Calculate Pressure at 1.00 km

Now we calculate \(P(h)\):\[P(1000) = 92 \, \text{atm} \times 0.942 \approx 86.6 \, \text{atm (Venus-atmospheres)}\]So, at 1.00 km above the surface, the pressure is approximately 86.6 Venus-atmospheres.
05

Calculate Root-Mean-Square Speed at the Surface

The root-mean-square speed \(v_{rms}\) is given by:\[v_{rms} = \sqrt{\frac{3RT}{M}}\]At the surface with \(T = 733.15\, K\) and \(M = 0.044\, kg/mol\):\[v_{rms \text{ at surface}} = \sqrt{\frac{3 \times 8.314 \times 733.15}{0.044}}\]Calculating gives:\[v_{rms \text{ at surface}} \approx 472.4 \, m/s\]
06

Calculate Root-Mean-Square Speed at 1.00 km

Since the temperature is constant with altitude, the root-mean-square speed will remain the same at 1.00 km altitude:\[v_{rms \text{ at 1.00 km}} = 472.4 \, m/s\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Greenhouse Effect
Venus is often cited as a prime example of the greenhouse effect. The greenhouse effect is essentially how Venus maintains its high surface temperatures. This occurs because the thick atmosphere of Venus, which is composed almost entirely of carbon dioxide (CO\(_2\)), traps heat. Solar radiation passes through the atmosphere and heats the planet's surface. However, when the surface emits infrared radiation (heat), this radiation is captured by the CO\(_2\) molecules. As a result, the heat is retained within the atmosphere, leading to higher temperatures.

This creates a runaway greenhouse effect, where heat builds up over time, making Venus far hotter than would be expected from solar heating alone. This is why Venus has an average surface temperature of about 460掳C, which is hotter than the surface of Mercury, even though Mercury is closer to the Sun.
Root-Mean-Square Speed
The concept of root-mean-square speed is crucial in understanding the behavior of gas molecules. It represents the average speed of gas particles in a given sample. The formula for the root-mean-square speed, \(v_{rms}\), is:
  • \(v_{rms} = \sqrt{\frac{3RT}{M}}\)
Where:
  • \(R\) is the universal gas constant, 8.314 J/(mol路K)
  • \(T\) is the temperature in Kelvin
  • \(M\) is the molar mass of the gas in kilograms per mole
At the surface of Venus, the temperature is about 733.15 K. Using the molar mass of CO\(_2\), which is 0.044 kg/mol, we calculate the root-mean-square speed of CO\(_2\) molecules at the surface. Importantly, because Venus's temperature is uniform with altitude, the \(v_{rms}\) is consistent up to 1.00 km above the surface.
Barometric Formula
The barometric formula is essential for calculating how atmospheric pressure decreases with increasing altitude. It is an expression that relates pressure to height in a planetary atmosphere, considering temperature and gravitational forces. The formula is:
  • \(P(h) = P_0 \, e^{-\frac{Mgh}{RT}}\)
Where:
  • \(P(h)\) is the pressure at height \(h\)
  • \(P_0\) is the surface pressure
  • \(M\) is the molar mass of the gas
  • \(g\) is the acceleration due to gravity
  • \(R\) is the universal gas constant
  • \(T\) is the temperature (constant here)
For Venus, with a given gravity and nearly constant temperature, we use this formula to find that the pressure at 1.00 km above the surface is about 86.6 Venus-atmospheres. This impressive pressure is due to the thick and heavy CO\(_2\) atmosphere of Venus.
Molar Mass of CO\(_2\)
Understanding the molar mass of carbon dioxide (CO\(_2\)) is important when analysing atmospheric conditions and behaviors. Molar mass is the mass of a given substance divided by the amount of substance (measured in moles), typically expressed in grams per mole. For CO\(_2\), it is 44.0 g/mol, or in terms of kilograms per mole, 0.044 kg/mol.

This molar mass is significant when calculating various atmospheric properties like pressure and speed of molecular motion. It influences the root-mean-square speed in the kinetic theory of gases and determines how gravity affects the gas, as seen in the barometric formula. The specific properties of CO\(_2\), including its molar mass, make it crucial in sustaining the dense and hot environment observed on Venus.

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Most popular questions from this chapter

We have two equal-size boxes, \(A\) and \(B\). Each box contains gas that behaves as an ideal gas. We insert a thermometer into each box and find that the gas in box \(A\) is at 50\(^\circ\)C while the gas in box \(B\) is at 10\(^\circ\)C. This is all we know about the gas in the boxes. Which of the following statements must be true? Which could be true? Explain your reasoning. (a) The pressure in \(A\) is higher than in \(B\). (b) There are more molecules in \(A\) than in \(B\). (c) A and B do not contain the same type of gas. (d) The molecules in \(A\) have more average kinetic energy per molecule than those in \(B\). (e) The molecules in \(A\) are moving faster than those in \(B\).

For diatomic carbon dioxide gas (CO\(_2\), molar mass 44.0 g/mol) at \(T\) = 300 K, calculate (a) the most probable speed \(\upsilon{_m}{_p}\); (b) the average speed \(\upsilon {_a}{_v}\); (c) the root-mean-square speed \(\upsilon {_r}{_m}{_s}\).

In an evacuated enclosure, a vertical cylindrical tank of diameter \(D\) is sealed by a 3.00-kg circular disk that can move up and down without friction. Beneath the disk is a quantity of ideal gas at temperature \(T\) in the cylinder (Fig. P18.50). Initially the disk is at rest at a distance of \(h\) = 4.00 m above the bottom of the tank. When a lead brick of mass 9.00 kg is gently placed on the disk, the disk moves downward. If the temperature of the gas is kept constant and no gas escapes from the tank, what distance above the bottom of the tank is the disk when it again comes to rest?

A person at rest inhales 0.50 L of air with each breath at a pressure of 1.00 atm and a temperature of 20.0\(^\circ\)C. The inhaled air is 21.0% oxygen. (a) How many oxygen molecules does this person inhale with each breath? (b) Suppose this person is now resting at an elevation of 2000 m but the temperature is still 20.0\(^\circ\)C. Assuming that the oxygen percentage and volume per inhalation are the same as stated above, how many oxygen molecules does this person now inhale with each breath? (c) Given that the body still requires the same number of oxygen molecules per second as at sea level to maintain its functions, explain why some people report 'shortness of breath' at high elevations.

At an altitude of 11,000 m (a typical cruising altitude for a jet airliner), the air temperature is -56.5\(^\circ\)C and the air density is 0.364 kg/m\(^3\). What is the pressure of the atmosphere at that altitude? (Note: The temperature at this altitude is not the same as at the surface of the earth, so the calculation of Example 18.4 in Section 18.1 doesn't apply.)
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