First slide

Change in internal energy

Question

Internal energy of an ideal gas is given by U=3PV. Then the value of $\frac{C_{p}}{C_{v}}$ of the gas is

Easy

A

$\frac{5}{3}$
B

$\frac{4}{3}$
C

$\frac{3}{2}$
D

$\frac{7}{5}$

Solution

$\begin{array}{l} U = 3 PV \Rightarrow U = 3 (nRT) \\ But U = n . C_{v} . T \Rightarrow {nC}_{v} T = 3 . nRT \Rightarrow C_{v} = 3 R \\ But C_{v} = \frac{R}{γ - 1} \Rightarrow \frac{R}{γ - 1} = 3 R \Rightarrow γ = \frac{4}{3} \end{array}$

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