The expansion of a gas follows the condition `pV^2`=constant .Show that such an expansion causes cooling of the gas.

Explanation

Assume that the initial volume of the gas is `V_1` at pressure `p_1` and at temperature `T_1`, after the expansion the corresponding quantities are `V_2,P_2` and `T_2` (say) . From ideal gas equation.
`(p_1V_1)/T_1=(p_2V_2)/T_2or(p_1V_1)/(p_2V_2)=T_1/T_2`
or,`(p_1V_1^2)/(p_2V_2^2)=(T_1V_1)/(T_2V_2)` [multiplying both sides by `V_1/V_2`]
As the condition is `p_1V_1^2=p_2V_2^2`, we have
`(V_1T_1)/(V_2T_2)=1or,T_1/T_2=V_2/V_1`
As the gas expands `V_2 gt V_1` therefore `T_1/T_2 gt 1` or , `T_2 lt T_1`.
`therefore` The gas cools down upon expansion.