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Chapter 1: Q. 1.18 (page 13)

Calculate the rms speed of a nitrogen molecule at room temperature.

Short Answer

Expert verified

Root mean square speed is 515.9m/s.

Step by step solution

01

Given information

Gas is Nitrogen.

Temp = room temp = 25oC= 298 K

02

Explanation

The root mean square velocity is calculated as

v=3kTm............................(1)

where

T = temperature

k = Boltzmann constant

m = mass

Mass of Nitrogen is calculated as

m=28u=28×1.6×10-27kg=4.65×10-26kg

Boltzmann constant is =1.38×10-23m2kgs-2K-1

Substitute the values in equation (1), we get

v=3×(1.38×10-23m2kgs-2K-1)×(298K)4.65×10-26kgv=515.9ms-1

Root mean square speed is 515.9m/s.

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Most popular questions from this chapter

Given an example to illustrate why you cannot accurately judge the temperature of an object by how hot or cold it feels to the touch?

Make a rough estimate of the total rate of conductive heat loss through the windows, walls, floor, and roof of a typical house in a cold climate. Then estimate the cost of replacing this lost energy over a month. If possible, compare your estimate to a real utility bill. (Utility companies measure electricity by the kilowatt-hour, a unit equal to MJ. In the United States, natural gas is billed in terms, where 1 therm = 105 Btu. Utility rates vary by region; I currently pay about 7 cents per kilowatt-hour for electricity and 50 cents per therm for natural gas.)

Calculate the rate of heat conduction through a layer of still air that is1mmthick, with an area of 1m2, for a temperature difference of 20∘C.

Even at low density, real gases don’t quite obey the ideal gas law. A systematic way to account for deviations from ideal behavior is the virial

expansion,

PV−nRT(1+B(T)(V/n)+C(T)(V/n)2+⋯)

where the functions B(T), C(T), and so on are called the virial coefficients. When the density of the gas is fairly low, so that the volume per mole is large, each term in the series is much smaller than the one before. In many situations, it’s sufficient to omit the third term and concentrate on the second, whose coefficient B(T)is called the second virial coefficient (the first coefficient is 1). Here are some measured values of the second virial coefficient for nitrogen (N2):

T(K)
B(cm3/mol)
100–160
200–35
300–4.2
4009.0
50016.9
60021.3
  1. For each temperature in the table, compute the second term in the virial equation, B(T)/(V/n), for nitrogen at atmospheric pressure. Discuss the validity of the ideal gas law under these conditions.
  2. Think about the forces between molecules, and explain why we might expect B(T)to be negative at low temperatures but positive at high temperatures.
  3. Any proposed relation between P, V, andT, like the ideal gas law or the virial equation, is called an equation of state. Another famous equation of state, which is qualitatively accurate even for dense fluids, is the van der Waals equation,
    (P+an2V2)(V−nb)=nRT
    where a and b are constants that depend on the type of gas. Calculate the second and third virial coefficients (Band C) for a gas obeying the van der Waals equation, in terms of aand b. (Hint: The binomial expansion says that (1+x)p≈1+px+12p(p−1)x2, provided that |px|≪1. Apply this approximation to the quantity [1−(nb/V)]−1.)
  4. Plot a graph of the van der Waals prediction for B(T), choosing aand bso as to approximately match the data given above for nitrogen. Discuss the accuracy of the van der Waals equation over this range of conditions. (The van der Waals equation is discussed much further in Section 5.3.)

List all the degrees of freedom, or as many as you can, for a molecule of water vapor. (Think carefully about the various ways in which the molecule can vibrate.)
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