Concept explainers
(a)
The rms speed of the
(a)
Explanation of Solution
Given:
The escape speed of the gas molecules in atmosphere of Mars is
Temperature is
Formula used:
Write the expression for the rms speed of the molecule.
Here,
Write the equation for the conversion of Celsius to kelvin.
Calculation:
Substitute
Substitute
Conclusion:
The rms speed of the
(b)
The rms speed of
(b)
Explanation of Solution
Given:
The escape speed of the gas molecules in atmosphere of Mars is
Temperature is
Formula used:
Write the expression for the rms speed of the molecule.
Calculation:
Substitute
Conclusion:
The root mean square speed for the oxygen molecule is
(c)
The root mean square speed of the
(c)
Explanation of Solution
Given:
The escape speed of the gas molecules in atmosphere of Mars is
Temperature is
Formula used:
Write the expression for the rms speed of the molecule.
Calculation:
Substitute
Conclusion:
The root mean square speed for the
(d)
Whether the hydrogen, oxygen and carbon dioxide molecules to be found in the atmosphere of the mars.
(d)
Explanation of Solution
Given:
The escape speed of the gas molecules in atmosphere of Mars is
Temperature is
Formula used:
Calculate
Here,
Calculation:
Substitute
Conclusion:
As
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Chapter 17 Solutions
Physics for Scientists and Engineers, Vol. 3
- Consider the Maxwell-Boltzmann distribution function plotted in Problem 28. For those parameters, determine the rms velocity and the most probable speed, as well as the values of f(v) for each of these values. Compare these values with the graph in Problem 28. 28. Plot the Maxwell-Boltzmann distribution function for a gas composed of nitrogen molecules (N2) at a temperature of 295 K. Identify the points on the curve that have a value of half the maximum value. Estimate these speeds, which represent the range of speeds most of the molecules are likely to have. The mass of a nitrogen molecule is 4.68 1026 kg. Equation 20.18 can be used to find the rms velocity given the temperature, Boltzmanns constant, and the mass of the atom or molecule. The mass of a nitrogen molecule is 4.68 1026 kg. vrms=3kBTm=3(1.381023J/K)4.681026kg=511m/s Using the results of Problem 28 and the rms velocity, we can calculate the value of f(v). f(vrms) = (3.11 108)(511)2 e(5.75106(511)2) = 0.00181 The most probable speed, for which this function has its maximum value, is given by Equation 20.20. vmp=2kBTm=2(1.381023J/K)(295K)4.681026kg=417m/s f(vmp) = (3.11108)(417)2 e(5.75106(417)2) = 0.00199 We plot these points on the speed distribution. The most probable speed is indeed at the peak of the distribution function. Since the function is not symmetric, the rms velocity is somewhat higher than the most probable speed. Figure P20.29ANSarrow_forwardFrom the MaxwellBoltzmann speed distribution, show that the most probable speed of a gas molecule is given by Equation 16.23. Note: The most probable speed corresponds to the point at which the slope of the speed distribution curve dNv/dv is zero.arrow_forwardCylinder A contains oxygen (O2) gas, and cylinder B contains nitrogen (N2) gas. If the molecules in the two cylinders have the same rms speeds, which of the following statements is false? (a) The two gases haw different temperatures. (b) The temperature of cylinder B is less than the temperature of cylinder A. (c) The temperature of cylinder B is greater than the temperature of cylinder A. (d) The average kinetic energy of the nitrogen molecules is less than the average kinetic energy of the oxygen molecules.arrow_forward
- Eight bumper cars, each with a mass of 322 kg. are running in a room 21.0 m long and 130 m wide. They have no driver, so they just bounce around on their own. The rms speed of the cars is 2.50 m/s. Repeating the arguments of Pressure, Temperature, and RMS Speed, find the average force per unit length (analogous to Pressure) that the cars exert the walls.arrow_forwardTwo monatomic ideal gases A and B are at the same temperature. If 1.0 g of gas A has the same internal energy as 0.10 g of gas B, what are (a) the ratio of the number of moles of each gas and (b) the ration of the atomic masses of the two gases?arrow_forwardSuppose that the rms speed of carbon dioxide molecules, with molar mass of 44.0 g/mol, in a flame is found to be 0.95 × 105 m/s. a)What temperature, in kelvins, does this represent?arrow_forward
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- At what temperature would the rms speed of the helium atoms (mass=6.66x10^-27kg) equal its escape speed from the earth which is 1.12x 10^4 m/s? Answer is given as 60509 K but cannot figure out how that could be it.arrow_forwardAt what temperature does the rms speed of (a) H2 (molecular hydrogen) and (b) O2 (molecular oxygen) equal the escape speed from Earth? At what temperature does the rms speed of (c) H2 and (d) O2 equal the escape speed from the Moon (where the gravitational acceleration at the surface has magnitude 0.16g)? Considering the answers to parts (a) and (b), should there be much (e) hydrogen and (f) oxygen high in Earth’s upper atmosphere, where the temperature is about 1000 K?arrow_forwardCalculate the root-mean-square (rms) speed of methane (CH4) gas molecules at a temperature of 325 K.arrow_forward
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