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Q. 1.23

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Found in: Page 17

### An Introduction to Thermal Physics

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

# 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

## 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

${U}_{\text{thermal}}=Nf\left(\frac{1}{2}kT\right)..............................\left(1\right)$

Where

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

${U}_{\text{thermal}}=3NkT\frac{1}{2}\phantom{\rule{0ex}{0ex}}{U}_{\text{thermal}}=\frac{3}{2}PV..............................\left(2\right)$

Substitute the values given we get

${U}_{\text{thermal}}=\frac{3}{2}PV\phantom{\rule{0ex}{0ex}}=\frac{3}{2}×\left(101325Pa\right)×\left(1×{10}^{-3}{m}^{3}\right)\phantom{\rule{0ex}{0ex}}{U}_{\text{thermal}}=151.98\mathrm{J}$

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

${U}_{\text{thermal}}=\frac{5}{2}PV\phantom{\rule{0ex}{0ex}}=\frac{5}{2}×\left(101325Pa\right)×\left(1×{10}^{-3}{m}^{3}\right)\phantom{\rule{0ex}{0ex}}{U}_{\text{thermal}}=253.31\mathrm{J}$

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