First slide

Change in internal energy

Question

The energy density U/V of an ideal mono-atomic gas is related to its pressure P as :

Moderate

A

$\frac{U}{V} = 3 P$
B

$\frac{U}{V} = \frac{3}{2} P$
C

$\frac{U}{V} = \frac{P}{3}$
D

$\frac{U}{V} = \frac{5}{2}$

Solution

$\begin{array}{l} I n t e r n a l E n e r g y U = n . C_{v} . T = n . \frac{3 R}{2} T = \frac{3}{2} P V \\ ∴ \frac{U}{V} = \frac{3}{2} P \end{array}$

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