Question

An ideal gas is expanding such that $P{{T}^{2}}=$constant. The coefficient of volume expansion of the gas is:
$\begin{align}
  & A.\text{ }\dfrac{1}{T} \\
 & B.\text{ }\dfrac{2}{T} \\
 & C.\text{ }\dfrac{3}{T} \\
 & D.\text{ }\dfrac{4}{T} \\
\end{align}$

Answer 1

Hint: We have provided an ideal gas which is expanding such that $P{{T}^{2}}=$ constant. Use the ideal gas equation and put the value of pressure in the given equation. Use expression of coefficient of volume expansion will give relation between volume, temperature, and coefficient. Differentiate this equation; put values rearrange it this way we get the answer.
Formula used:
Expression of volume expansion is given by,
$V={{V}_{0}}(1+\gamma t)$
Where, V is the volume at temperature t
 ${{V}_{0}}$ Is the volume at temperature t = 0
$\gamma $ Coefficient of volume expansion
t is temperature.

Complete answer:
We have given that ideal gas is expanding such that $P{{T}^{2}}=$ constant….. (1)
According to ideal gas equation,
$\begin{align}
  & PV=nRT \\
 & \dfrac{PV}{T}=\text{constant} \\
 & P=\dfrac{\text{constant}\times T}{V}......\left( 2 \right) \\
\end{align}$
We need to calculate the values of coefficient of volume expansion of gas denoted by γ
Coefficient of volume expansion of gas is given by
$V={{V}_{0}}(1+\gamma t).......\left( 3 \right)$
Now put equation (2) in equation (1), we get
$\begin{align}
  & \dfrac{\text{constant}\times T}{V}\times {{T}^{2}}=\text{constant} \\
 & \Rightarrow \dfrac{{{T}^{2}}}{V}=\text{constant=k} \\
 & \Rightarrow V=\dfrac{1}{k}{{T}^{3}}={{k}^{1}}{{T}^{3}}.....\left( 4 \right)\text{ }\left( \because {{k}^{1}}=\dfrac{1}{k} \right) \\
\end{align}$
Now differentiate equation (3) w.r.to t, we get
$\begin{align}
  & \dfrac{dV}{dt}={{V}_{0}}(\gamma .1) \\
 & \dfrac{dV}{dt}={{V}_{0}}\gamma .......\left( 5 \right) \\
\end{align}$
Now differentiate equation (2) w.r.to T, we get.
$\dfrac{dV}{dT}={{k}^{1}}3{{T}^{2}}$
Put values of ${{k}^{1}}$ from equation (4) we get.
$\begin{align}
  & \dfrac{dV}{dT}=\dfrac{V}{{{T}^{3}}}\times 3{{T}^{2}} \\
 & \dfrac{dV}{dT}=\dfrac{3V}{T} \\
 & \therefore \dfrac{dV}{V}=3\dfrac{dT}{T} \\
\end{align}$
Rearrange we get,
$dV=\dfrac{3}{T}VdT......(6)$
We know that original expression of volume expansion is given by,
$\Delta V=\gamma {{V}_{0}}\Delta T......\left( 7 \right)$
Compare (6) and (7) we get,
$\gamma =\dfrac{3}{T}$
Hence the coefficient of volume expansion of the gas is $\dfrac{3}{T}$.

Therefore the correct option is (c).

Note:
Coefficient of volume expansion depends on the nature of the material. Like volume expansion we have linear expansion. Volume expansion is applicable for gas whereas linear expansion is applicable for solid coefficient of linear expansion is denoted by ∝. The term increases in volume (V) depends on original volume (${{V}_{0}}$) and rise in temperature (t).