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Q. 1.23

An Introduction to Thermal Physics
Found in: Page 17
An Introduction to Thermal Physics

An Introduction to Thermal Physics

Book edition 1st
Author(s) Daniel V. Schroeder
Pages 356 pages
ISBN 9780201380279

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Short Answer

Calculate the total thermal energy in a liter of helium at room temperature and atmospheric pressure. Then repeat the calculation for a liter of air.

Total thermal energy in a liter of helium : 151.98 J

Total thermal energy in a liter of air : 253.31.J

See the step by step solution

Step by Step Solution

Step1: Given information

Temperature T = 298 K

Degree of freedom of Helium is f=3

Step2: Explanation

Total thermal energy is for a system contains N molecules each with f degree of freedom is given as

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


N = number pf molecules

k = Boltzmann constants

f = degree of freedom

T = Temperature

A liter of helium at room temperature and pressure of atm = 101325 Pa

Using ideal gas law P V=n k T, Find the thermal energy of

Uthermal =3NkT12Uthermal =32PV..............................(2)

Substitute the values given we get

Uthermal =32PV=32×(101325 Pa)×(1×10-3m3)Uthermal =151.98 J

For air , Number of degree of freedom is f=5
Find the thermal energy by substituting values in equation 2.

Uthermal =52PV=52×(101325Pa)×(1×10-3m3)Uthermal =253.31 J

Most popular questions for Physics Textbooks

Measured heat capacities of solids and liquids are almost always at constant pressure, not constant volume. To see why, estimate the pressure needed to keep V fixed as T increases, as follows.

(a) First imagine slightly increasing the temperature of a material at constant pressure. Write the change in volume,dV1, in terms of d T and the thermal expansion coefficient β introduced in Problem 1.7.

(b) Now imagine slightly compressing the material, holding its temperature fixed. Write the change in volume for this process, d V2, in terms of d P and the isothermal compressibility κT, defined as


(c) Finally, imagine that you compress the material just enough in part (b) to offset the expansion in part (a). Then the ratio of dP to dT is equal to (P/T)V, since there is no net change in volume. Express this partial derivative in terms of β and κT. Then express it more abstractly in terms of the partial derivatives used to define β and κT. For the second expression you should obtain


This result is actually a purely mathematical relation, true for any three quantities that are related in such a way that any two determine the third.

(d) Compute β,κT, and (P/T)V for an ideal gas, and check that the three expressions satisfy the identity you found in part (c).

(e) For water at 25C,β=2.57×104K1 and κT=4.52×1010Pa1. Suppose you increase the temperature of some water from 20C to 30C. How much pressure must you apply to prevent it from expanding? Repeat the calculation for mercury, for which (at 25C ) β=1.81×104K1 andκT=4.04×1011Pa1

Given the choice, would you rather measure the heat capacities of these substances at constant v or at constant p ?


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