Question

The pressure $(P)$ and temperature $(T)$ relationship of an ideal gas obeys the equation $P{T}^2=$ constant. The volume expansion coefficient of the gas will be :

• A. $3T^2$
• B. $\frac{3}{T^2}$
• C. $\frac{3}{T^3}$
• D. $\frac{3}{T}$

Answer 1

Answer: (D)

$P{T}^2=$ constant, Using
$PV=nRT$
$P=\frac{nRT}{V}$
$P{T}^2=\frac{nRT}{V}\times T^2=\text{ constant }$
$\Rightarrow T^3=KV$
$\text{ So, }\frac{d}{dT}(KV)=3T^2$
$\Rightarrow \frac{KdV}{dT}=3T^2$
$\Rightarrow dV=\frac{3T^2}{K}dT$
$dV=V\gamma dT$
$\Rightarrow \gamma V=\frac{3T^2}{K}$
$\Rightarrow \gamma =\frac{3T^2}{KV}=\frac{3T^2}{T^3}=\frac{3}{T}$