Find the minimum attainable pressure of one mole of an ideal gas. If during its expansion its temperature and volume are related as `T=T_(0)+alphaV^(2)` where `T_(0)&alpha` are positive constants

Explanation

Given that one mole of gas is used, thus from gas law, we have
`PV=RT`
or `" "P=(RT)/(V)=( R )/(V)(T_(0)+alpha V^(2))`
`["As "T=T_(0)+alpha V^(2)]`
Here pressure P will be minimum when
`(dP)/(dV)=0`
or `" "(dP)/(dV)=-(RT_(0))/(V^(2))+alphaR=0`
or `" "V=sqrt((T_(0))/(alpha))`
Thus pressure of gas is minimum when its volume is `V=sqrt((T_(0))/(alpha))` and at this volume its temperature is given as
`T=T_(0)+alpha V^(2)`
`T=T_(0)+alpha(sqrt((T_(0))/(alpha)))^(2)=2T_(0)`
Thus minimum value of pressure is
`P_("min")=(RT)/(V)=(R(2T_(0)))/(sqrt((T_(0))/(alpha)))=2Rsqrt(T_(0)alpha)`