An ideal gas is expanding such that $p T^2=$ constant. The coefficient of volume expansion of the gas is

An ideal gas is expanding such that $p T^2=$ constant. The coefficient of volume expansion of the gas is
  1. $\frac{1}{T}$
  2. $\frac{2}{T}$
  3. $\frac{3}{T}$
  4. $\frac{4}{T}$

Solution

$p T^2=$ constant $\begin{array}{ll}\therefore \quad & \left(\frac{n R T}{V}\right) T^2=\text { constant } \\ \text { or } \quad T^3 V^{-1}=\text { constant }\end{array}$ Differentiating the equation, we get $ \frac{3 T^2}{V} \cdot d T-\frac{T^3}{V^2} \cdot d V=0 $ or $ 3 \cdot d T=\frac{T}{V} \cdot d V $ From the equation, $d V=V \gamma \cdot d T$ $\gamma$ = coefficient of volume expansion of gas $=\frac{d V}{V \cdot d T}$ From Eq. (i) $\gamma=\frac{d V}{V \cdot d T}=\frac{3}{T}$ $\therefore$ correct answer is (c)

Asked in: JEE Advanced 2008 (Paper 1)

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