Question

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

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

Answer 1

Answer: (D)

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