Calculate the total thermal energy in a gram of lead at room temperature, assuming that none of the degrees of freedom are "frozen out" (this happens to be a good assumption in this case).

Short Answer

Expert verified

Total thermal energy is 35.3 J.

Step by step solution

01

Given information

Temperature T = 298 K

Number of degree of freedom f= 6 (3 kinetic + 3 potential)

02

Explanation

Total thermal energy is calculated for system is

Uthermal=Nf12kT......................(1)

Where

N = number of molecules

k = Boltzmann constants

T = temperature

f = number of degree of freedom

lead has degree of freedom of f= 6

molecular mass of lead is 207.2 u

we can calculate number of molecules as

N=massmolecular mass

Here mass is 1 gram =1 x 103kg

molecular mass is 207.2 u = 207.2 x 1.6 x 10-27 kg

Substitute the values, we get

N=1×10-3k207.2×1.6×10-27kgN=2.91×1021

Now substitute the values in equation (1) to get thermal energy

Uthermal=62×(2.91×1021)×(1.38×10-23JK-1)×(298K)Uthermal=35.3J

Total thermal energy is 35.3 J.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

When the temperature of liquid mercury increases by one degree Celsius (or one kelvin), its volume increases by one part in 550,000 . The fractional increase in volume per unit change in temperature (when the pressure is held fixed) is called the thermal expansion coefficient, β :
βΔV/VΔT
(where V is volume, T is temperature, and Δ signifies a change, which in this case should really be infinitesimal if β is to be well defined). So for mercury, β =1 / 550,000 K-1=1.81 x 10-4 K-1. (The exact value varies with temperature, but between 0oC and 200oC the variation is less than 1 %.)
(a) Get a mercury thermometer, estimate the size of the bulb at the bottom, and then estimate what the inside diameter of the tube has to be in order for the thermometer to work as required. Assume that the thermal expansion of the glass is negligible.
(b) The thermal expansion coefficient of water varies significantly with temperature: It is 7.5 x 10 -4 K-1 at 100oC, but decreases as the temperature is lowered until it becomes zero at 4oC. Below 4oC it is slightly negative, reaching a value of -0.68 x 10-4K-1 at 0oC. (This behavior is related to the fact that ice is less dense than water.) With this behavior in mind, imagine the process of a lake freezing over, and discuss in some detail how this process would be different if the thermal expansion coefficient of water were always positive.


At the back of this book is a table of thermodynamic data for selected substances at room temperature. Browse through the CPvalues in this table, and check that you can account for most of them (approximately) using the equipartition theorem. Which values seem anomalous?

Make a rough estimate of the thermal conductivity of helium at room temperature. Discuss your result, explaining why it differs from the value for air.

A mole is approximately the number of protons in a gram of protons. The mass of a neutron is about the same as the mass of a proton, while the mass of an electron is usually negligible in comparison, so if you know the total number of protons and neutrons in a molecule(i,e., its "atomic mass"), you know the approximate mass(in grams) of a mole of these molecules. Referring to the periodic table at the back of this book ,find the mass of a mole of each of the following : Water, nitrogen (N2), lead, quartz (Si O2)

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,

PVnRT(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)(Vnb)=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)p1+px+12p(p1)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.)
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free