ABC26PH0223 · Thermodynamics
Subject: Physical Chemistry · Chapter: Thermodynamics · Topic: Thermodynamics – General · Exam: CSIR-NET JUNE 2011 · Marks: 2 · Difficulty: Medium
$\begin{array}{l}1.1 2.2 3.3 4.4\end{array}$
(a)mkT
(b)AkT
(c)kT/m
(d)kT/A
Answer
Explanation
()² ,+ Average energy per particle E= ………………..1 + ,- .- / (Where f= single particle, partition function,) here, f= V 0 𝑑𝑓 𝑑 𝐴𝑇 5 = 𝑑𝑇 𝑑𝑇 𝑉 ,+ . = 𝑚𝑇 589 ……………..2 ,- 0 Put this value in equation (1), we get :- ; . E= <=/ 𝑚𝑇 589 = 𝑚𝑘𝑇 0 >
Explanation as extracted from the printed page; notation may be imperfect.
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